## Guidelines/Expectatations

- Each student is expected to make at least three contributions to the wiki before each quiz (that is, in the two weeks in between quizzes.)
- A contribution may consist of a worked solution or an edit of a solution posted by another student. You can choose from among any of the exercises listed in the daily assignment.
- To receive credit for a worked solution, you should post the original problem as well as a complete solution for the problem. (It's ok if your work has an error or can be improved. However, what you post should be what you believe is a complete solution.)
- To receive credit for an edit, you should substantially improve the original solution. This can include correcting errors, filling in gaps, or polishing writing.

- All work posted on the wiki should meet the guidelines for good mathematical writing, including grammatically correct writing and appropriate use of (Latex'ed) symbols.
- To receive credit for wiki contributions, you should keep track of your work on your student page.
- See the Writing Guidelines for more details about writing expectations and the Participation Guidelines for more details about how your wiki contributions fit into your participation grade.
- Some of the suggested exercises may contain partial solutions in the back of the textbook. To receive credit for a solution, your work must go beyond what's found in the textbook. Copying a solution from another student or an outside source is not acceptable. (See the syllabus for more information on academic honesty.)

### Feedback on Daily Exercises

The daily exercises are intended to give you a chance to try out ideas and critically reflect on both your own work and your classmates' work. This means that I want to give you a chance to work things out on your own, but also get some basic feedback from me. With this in mind, I'll offer the following general critiques of your work. (These are good starting points for revisions!)

- Gap/Error: This means that there is either an error in the reasoning or calculation, or there is a significant gap in the explanation.
- Confusing/Ambiguous: This means that I find your reasoning/calculation difficult to follow or there are aspects of it that are not clearly explained.
- Formatting/Writing: There are problems with the presentation of your ideas. This includes issues with your use of notation, grammatical or spelling errors, the structure of sentences, or the general organization.
- $\checkmark$: The solution looks mathematically correct and complete and is well-written.

## Exercises

**0.00**: Here is a template. For example, I can represent the real numbers as $\bf R$. In general, include the original problem here.

Put your solution here.

**0.8**: Suppose $a$ and $b$ are integers that divide the integer $c$. If $a$ and $b$ are relatively prime, show that $ab$ divides $c$. Show, by example, that if $a$ and $b$ are not relatively prime, then $ab$ need not divide $c$.

Because $a$ and $b$ divide $c$, then by Theorem 0.1, $c=aq_1$ and $c=bq_2$ where $q_1$ and $q_2$ are integers. We also know that $a$ and $b$ are relatively prime, so by the corollary to Theorem 0.2, there are integers $s$ and $t$ such that $1=as+bt$. We multiply this equation by $c$ to get $c=cas+cbt$. Because $c=aq_1$ and $c=bq_2$, we substitute the first $c$ on the right side of the equation with $bq_2$ and the second $c$ on the right side of the equation with $aq_1$, giving $c=absq_2+absq_1$. The right side of this equation is divisible by $ab$, so the left side of the equation is divisible by $ab$, and therefore $c$ is divisible by $ab$.

If $a$ and $b$ are not relatively prime, this is not necessarily true. For instance, let $a=5, b=10, c=20$. Then $a$ divides $c$ because 20/5=4, $b$ divides $c$ because 20/10=2, but $ab$ does not divide $c$ because 20/50 is not an integer.

Feedback: $\checkmark$

**0.22**: For every positive integer $n$, prove that $1+2+ \cdots +n=\frac{n(n+1)}{2}.$

Proof by induction.

Show $n=1$ is a solution.

(1)Then show $n$ being a solution implies $n+1$ is a solution. In other words, show $1+2+ \cdots +(n+1) = \frac{(n+1)(n+2)}{2}$ by assuming $1+2+ \cdots +n=\frac{n(n+1)}{2}.$

(2)Therefore, by induction, for all positive integer $n$, $1+2+ \cdots +n=\frac{n(n+1)}{2}.$

Feedback: $\checkmark$

**1.6**: In $D_n$, explain geometrically why a reflection followed by a reflection must be a rotation.

Let $A$ represent the initial visible, face up surface of an object in $D_n$. Let $B$ represent the surface opposite of $A$.

In $D_n$, a reflection will always result in surface $B$ being face up. A second reflection will flip the object again, resulting in surface $A$ being face up. After two reflections, the object may be in a different position, but the initial surface is the visible one. Since the object was reflected twice, it is possible to return the object to it's identity through rotations.

If two reflections of the same type were performed one after another, no rotation would be necessary to return the object to it's identity.

Feedback: $\checkmark$. (Note: The identity can be thought of at the 0-rotation.)

**1.20**: Bottle caps that are pried off typically have 22 ridges around the rim. Find the symmetry group of such a cap.

For now, assume a bottle cap is a flat, 2 dimensional object where each side is exactly the same. Also assume there in no logo or other markings.

This bottle cap has exactly 22 lines of reflection. 11 of which are formed by making a line from one ridge to the opposite side ridge. The other 11 are formed by making the line from the gap between any two ridges to the gap on the exact opposite side. There are only 11 of each because any reflection line could be formed by starting at either ridge or gap in the line, dividing the possibilities by 2 from the 22 ridges and 22 gaps.

This bottle cap also has rotational symmetry, one for each of the 22 vertices. By that, we mean that for any ridge you start at, you can rotate it to any of the 22 ridges, including itself with the identity rotation. This shape has the rotational symmetries of $R$_{22$n$/360}; where $n$ is an integer so that $0 \leq n \leq 22$

With 22 reflection lines and 22 rotations, this bottle cap has the same symmetries as a 22-gon, which has a symmetry group of $D$_{22}.

Now, of course, real bottle caps aren't flat objects, they have two distinct sides, sharp side up and sharp side down. Let's first look at sharp side up. It's clear that all the rotational symmetry will hold, but reflections seem to be lost because flipping the cap will make the bottle cap sharp side down. But, we must remember that a reflection doesn't actually flip the object, it more like how you'd see the object in a mirror, switching the two sides of the reflection line. With this interpretation it's clear that all the reflections do hold. Similarly, all the rotations and reflections hold when it's sharp side down. Since all the symmetries are still there, the bottle cap really does have symmetry $D$_{22}.

Of course, if the bottle cap is bent or has any other markings or imperfections, the bottle cap will lose some, of not all, of it's symmetries depending on the imperfection.

Old (before 10/20/2013) Feedback: Confusing/Ambiguous. (Are the two sides of the bottle cap distinct?)

**2.03**: Show that {1,2,3} under multiplication modulo 4 is not a group but that {1,2,3,4} under modulo 5 is a group.

1 | 2 | 3 | |
---|---|---|---|

1 | 1 | 2 | 3 |

2 | 2 | 0 | 2 |

3 | 3 | 2 | 1 |

This is not a group. One reason for this is that the element $2$ does not have an inverse. Since the identity is 1, in order for $2$ to have an inverse, $2b = b2 = 1$. Since there is no element $b$ such that $2b = 1$, {1,2,3} under multiplication modulo 4 is not a group.

Cayley Table for {1,2,3,4} under multiplication modulo 5:1 | 2 | 3 | 4 | |
---|---|---|---|---|

1 | 1 | 2 | 3 | 4 |

2 | 2 | 4 | 1 | 3 |

3 | 3 | 1 | 4 | 2 |

4 | 4 | 3 | 2 | 1 |

This is a group. It is closed, the identity is 1, the inverse of 1 is 1, 2 is 3, 3 is 2, and 4 is 4, and it is associative ($(ab)c = a(bc)$ is true for all elements).

**2.06**: Give an example of group elements $a$ and $b$ with the property that $a^{-1}$$ba \neq b$

Using the Cayley Table for a triangle, we can use the group as our Cayley Table.

We can chose $a = R_{240}$ and $b = V$. ($R_{240}$ being a rotation of the triangle of 240 $^\circ$ and [$V$ being a vertical reflection)

Then, our $a^{-1} = R_{120}$ by using the Cayley Table to find our inverse.

We want to prove that $a^{-1}$$ba \neq b$ or $R_{120} V R_{240} \neq V$

By using the Cayley Table or the triangle itself to do these symmetries, we get that $D \neq V$ ($D$ being a diagonal reflection)

Therefore, $a^{-1}$$ba \neq b$

Feedback: $\checkmark$

**2.22**: Give an example of a group with 105 elements. Give two examples of groups with 44 elements.

According to the text, "The set $Z$_{$n$} = {0,1,…,$n$-1} for $n \geq 1$ is a group under addition modulo $n$." $Z$_{$n$} can take any positive integer $n$ and create a group with $n$ elements.

$Z$_{105} is an example of a group with 105 elements.

$Z$_{44} is an example of a group with 44 elements.

Another group with 44 elements is $D$_{22} (See 1.20).

**2.25**: Suppose the table below is a group table. Fill in the blank entries.

e | a | b | c | d | |
---|---|---|---|---|---|

e | e | - | - | - | - |

a | - | b | - | - | e |

b | - | c | d | e | - |

c | - | d | - | a | b |

d | - | - | - | - | - |

**Identity**There is an element e (called the identity) in $G$ such that $ae=ea=a$ for all $a$ in $G$. Therefore, it is true that for any element, $x$ in the $e$ row and column, the result is $x$.

Therefore,

e | a | b | c | d | |
---|---|---|---|---|---|

e | e | a | b | c | d |

a | a | b | - | - | e |

b | b | c | d | e | - |

c | c | d | - | a | b |

d | d | - | - | - | - |

**Inverses**For each element $a$ in $G$, there is an element $b$ in $G$ (called the inverse of $a$) such that $ab = ba = e$. Therefore since $ad = e$, $da = e$ and since $bc = e$, $cb = e$.

e | a | b | c | d | |
---|---|---|---|---|---|

e | e | a | b | c | d |

a | a | b | - | - | e |

b | b | c | d | e | - |

c | c | d | e | a | b |

d | d | e | - | - | - |

**Associativity** The operation is associative; that is, $a(bc) = (ab)c$ for all $a, b, c$ in $G$.

Therefore, the following is true:

$c(cb) = (cc)b$

$ce = ab$

$c =$ **c**

$c(ab) = (ca)b$

$cc = db$

$a =$ **a**

$d(bc) = (db)c$

$de = ac$

$d =$ **d**

$a(ad) = (aa)d$

$ae = bd$

$a =$ **a**

$a(cc) = (ac)c$

$aa = dc$

$b =$ **b**

$a(cd) = (ac)d$

$ab = dd$

$c =$**c**

Therefore,

e | a | b | c | d | |
---|---|---|---|---|---|

e | e | a | b | c | d |

a | a | b | c | d | e |

b | b | c | d | e | a |

c | c | d | e | a | b |

d | d | e | a | b | c |

**2.26**: Prove that if $(ab)^2 = a^2b^2$ in a group G, then $ab=ba$

Assume $(ab)^2 = a^2b^2$

then $(ab) (ab) = a \centerdot a \centerdot b \centerdot b$

$a(ba)b = a(ab)b$

Then, we multiply each side by the inverses of $a$ and $b$, $a^{-1}$ and $b^{-1}$ to get

$ab=ba$

Feedback: Confusing/Ambiguous. (Make more clear how you are using each group property.)

$\textbf{EDIT:}$

Assume $(ab)^2 = a^2b^2$

then $(ab) (ab) = a \centerdot a \centerdot b \centerdot b$

$a(ba)b = a(ab)b$ by associativity of the group.

Then, because a and b are group elements so the both have an inverse, we multiply each side by the inverses of $a$ and $b$, $a^{-1}$ and $b^{-1}$ to get

$a(ba)b=a(ab)b$=$a^{-1}a(ba)bb^{-1}=a^{-1}a(ab)bb^{-1}$=$ba=ab$, which is $ab=ba$

$$

**2.28**: Prove that the set of all rational numbers of the form $3^m 6^n$, where *m* and *n* are integers, is a group under multiplication.

As we know, proving a group (which we will call *G*) requires proving it has an identity, an inverse for each element, associativity for the operation, and closure under the group. Note that all numbers that can be written as $3^m 6^n \in G$, where $m, n \in \mathbb{Z}$, are rational, so we are considering the elements containing any *m* and *n* to be in the group.

**Identity:** The multiplicative identity 1 can be given when $m = n = 0$. The identity is $3^0 6^0$.

**Inverse:** To find the general inverse $3^s 6^t$ where $s, t \in \mathbb{Z}$ we use the fact that:

Without getting into the complications of logarithms, one can see that a solution to this equivalence is that $s = -m, t = -n$, which are still integers. The inverse for $3^m 6^n$ is $3^{-m} 6^{-n}$.

**Associativity:** This means that $a(bc) = (ab)c$ where $a, b, c \in G$. If we let $i, j, m, n, s, t \in \mathbb{Z}$, the following shows this property:

Thus this group is associative.

**Closure:** For $m, n, s, t \in \mathbb{Z}$, we see that:

If $m, n, s, t \in \mathbb{Z}$, then $m + s, n + t \in \mathbb{Z}$ as well. We can apply this infinitely and always get an element that is in group *G*, thus it is closed.

These four show that *G* is in fact a group, based on our definition of a group.

Feedback: $\checkmark$

**2.33**: Suppose that $G$ is a group with the property that for every choice of elements in $G$, $axb=cxd$ implies $ab=cd$. Prove that $G$ is Abelian. ("Mixed cancellation" implies commutativity.)

Let $a \in G$ and $b \in G$, let $e\in G$ by the identity element, and let $a^{-1}$ be the inverse of $a$. We start with the obviously true statement that $b=b$. By the definition of an identity, $eb=be=b$. So we substitute on both sides of the $b=b$ equation to get $eb=be$. Then by the definition of an inverse, $aa^{-1}=a^{-1}a=e$. We substitute into $eb=be$ to get $aa^{-1}b=ba^{-1}a$. Finally, we use the middle cancellation property of the group that we are given to obtain $ab=ba$. This is the definition of an Abelian group. Therefore, middle cancellation implies the group is Abelian.

Feedback: $\checkmark$

**3.8**: Let $x$ belong to a group. If $x$^{2}$\neq e$ and $x$^{6}$= e$, prove that $x$^{4}$\neq e$ and $x$^{5}$\neq e$. What can we say about the order of $x$?

According to the theorem of identity, if $x$^{$n$}$=e$, then $x$^{$2n$}=($x$^{$n$})($x$^{$n$})$=e=x$^{$n$}.

Which means, every multiple of the order of an element can be used as a power of $x$ to equal the identity.

$x$^{6} does not necessarily mean that |$x$|=6, but rather 6 is a multiple of the order.

The order can either be 1, 2, 3, or 6.

Since $x$^{2}$\neq e$, 2 cannot be the order. Neither can 1, which is divisible by 2.

This proves that $x$^{5}$\neq e$, since 5 is a multiple of 1.

This also proves $x$^{4}$\neq e$, since 4 is a multiple of 2.

The only thing we can say about the order of $x$ is that it is either 3 or 6.

$\textbf{EDIT:}$

By Corollary 2 under Theorem 4.2, when G is a group and x is an element of order $n$ in G, if $x^k = e$, then $n$ divides $k$. For this problem, it is given that $x^6 = e$, so for use with this corollary, $k$ = 6. Then, n divides 6, or 6/$n$ must be true. Therefore, 3 and 6 are our only choices for $n$ as stated above since we want our $n$ values to divide into 6 evenly so that the corollary is true.

Feedback: Gap/Error. (How do you know that only multiples work?)

**3.9:** Show that if *a* is an element of a group *G*, then $|a| \leq |G|$.

We could choose to break up the proof into whether the group *G* and element *a* have an infinite or finite order, but I feel that this complicates the explanation and can be avoided. First, we note that as $a \in G$, *G* is a nonempty group and thus $|G| > 0$. It's also true that $|a| > 0$, which is an intrinsic property of group elements as $|a| = 1$ is the smallest positive integer that any element can have such that $a^{|a|} = e$.

Note that according to Theorem 4.1: Corollary 1, $|a| = |<a>|$ for any group element *a*.

Regardless of the finitude of orders, we can see that assuming $|a| > |G|$ implies that there are at least $|a| - |G|$ elements of $<a>$ that are not elements of *G*, so that *a* is not closed under the operation of *G*, which is a contradiction. So it's impossible that $|a| > |G|$.

For sake of argument, let's note that $G = <a>$ implies the case that $|a| = |G|$.

Also that if $G = \{<a>, b, \cdots\}$, where $b \notin <a>$, then $|a| < |G|$, as there are more elements in *G* than in $<a>$.

Thus, for $a \in G$, we have shown that $0 < |a| \leq |G|$.

Feedback: Gap/Error. (You are assuming that $|a|=|<a>|$?)

**3.10**: Show that U(14) = <3> = <5>. [Hence, U(14) is cyclic.] Is U(14) = <11>?

U(14) = {1,3,5,9,11,13}

<3> = {3, 3^{2}, 3^{3}, 3^{4},…} = {3,9,13,11,5,1,3,9,13,…}

<5> = {5,11,13,9,3,1,5}

Since <3> and <5> produce the same elements that make up U(14), we can say that U(14) = <3> = <5> and therefore 3 and 5 are generators for U(14) and U(14) is cyclic.

<11> = {11,9,1,11,9,…} It does not produce all the elements in U(14) so they are not equal.

Feedback: $\checkmark$ (Note that we normally don't repeat elements when listing set elements.)

**3.14**: Suppose that H is a proper subgroup of Z under addition and H contains 18,30,40. Determine H.

18,30,and 40 are in the set.

To start, we need numbers to add to 18 to get 30 and 40, which are 12 and 22.

(12,18,22,30,40)

Then we need a number to add to 12 to get 18, which is 6.

(6,12,18,22,30,40)

Next we find a number to add to 18 to get 22, which is 4

(4,6,12,18,22,30,40)

Now we can see that we need to add 2 to get from 4 to 6,

(2,4,6,12,18,22,30,40), and because 2 is in the group, and there are no odd integers, the group H is equal to all even integers.

Feedback: Formatting/Writing, Confusing/Ambiguous. (How are the properties of a subgroup being used?)

**EDIT:**

$H$ is equal to all even integers. This includes 0, which serves as the identity of a group under addition.

Since $a+(-a)=0$ for any $a\in H$, the inverse of $a$ is $-a$.

Take elements $a,b,c\in Z$.

$a(bc)=a+(b+c)=a+b+c=(a+b)+c=(ab)c$

Since addition between integers is associative, then $H$ is associative.

$H$ is a subgroup of $Z$.

**3.17**: For each divisor k>1 of n, let $U_k(n)={{x \in U(n)|x mod k =1}}$. List the elements of $U_4(20), U_5(20), U_5(30), and U_10(30)$. Prove that $U_k(n)$ is a subgroup of U(n). Let $H={{x \in U(10)|x mod 3 =1}}$. Is H a subgroup of U(10)?

$U_4(20) = {1, 9, 13, 17}$

$U_5(20) = {1, 11}$

$U_5(30) = {1, 11}$

$U_10(30) = {1, 11}$

We can prove $U_k(n)$ is a subgroup of U(n) if we can prove its closed. We know $U_k(n)$ closed because (ab) mod k (a mod k)(b mod k) = 1*1 = 1.

H is not a subgroup of U(10) because H is not closed.

**Daily Assignment 9/10 $U(8)$**: For the value n=8, determine

-the elements of U(n)

-the order of the group U(n) and the order of each of its elements

-Try to find at least one subgroup

$U(8)=\{1,3,5,7\}$

1 | 3 | 5 | 7 | |
---|---|---|---|---|

1 | 1 | 3 | 5 | 7 |

3 | 3 | 1 | 7 | 5 |

5 | 5 | 7 | 1 | 3 |

7 | 7 | 5 | 3 | 1 |

$\mid U(8) \mid = 4$

Order of elements:

$\mid 1 \mid = 1$

$\mid 3 \mid = 2$

$\mid 5 \mid = 2$

$\mid 7 \mid = 2$

One possible subgroup: $\{1,3\}$

In order for $\{1,3\}$ to be a subgroup it must be closed under multiplication, be associative, have an identity, and inverses.

Closure: $\{1,3\}$ is closed under multiplication. When any element of the subgroup is multiplied by itself or another subgroup element, you will always get an element of the subgroup as an answer.

Associativity: $\{1,3\}$ is associative. (1)(3)=(3)(1)

Identity: The identity for $\{1,3\}$ is 1

Inverses: Every element of the subgroup $\{1,3\}$ has an inverse that is an element in the subgroup.

inverse of 1 = 1

inverse of 3 = 3

Therefore, $\{1,3\}$ is one possible subgroup of $U(8)=\{1,3,5,7\}$

Feedback: Gap/Error. (Associativity is not the same as commutativity.)

$\textbf{EDIT:}$

The definition of associativity is that for all $a,b,c$ in the set, $(ab)c=a(bc)$. But the definition does not say [[a,b,c]] must be distinct. So we can say $\{1,3\}$ is associative because $1(31)=(13)1$,$1(33)=(13)3$,$3(31)=(33)1$,$1(11)=(11)1$, $3(33)=(33)3$, $3(11)=(31)1$, $1(13)=(11)3$, $3(13)=(31)3$.

**Daily Assignment 9/10 $U(5)$**: For the value n=5, determine

-the elements of U(n)

-the order of the group U(n) and the order of each of its elements

-Try to find at least one subgroup

$U(5)=\{1,2,3,4\}$

1 | 2 | 3 | 4 | |
---|---|---|---|---|

1 | 1 | 2 | 3 | 4 |

2 | 2 | 4 | 1 | 3 |

3 | 3 | 1 | 4 | 2 |

4 | 4 | 3 | 2 | 1 |

$\mid U(5) \mid = 4$

Order of elements:

$\mid 1 \mid = 1$

$\mid 2 \mid = 4$

$\mid 3 \mid = 4$

$\mid 4 \mid = 2$

We know $\{1,4\}$ is a subgroup if it is closed under multiplication, is associative, has an identity, and has inverses.

We know $\{1,4\}$ is closed under multiplication because if you multiply any element of the subgroup by another element of the subgroup you will get an element of the subgroup. We know $\{1,4\}$ is associative because (1)(4)=(4)(1).

We know the identity for $\{1,4\}$ is 1 and we know every element of the subgroup $\{1,4\}$ has an inverse that is an element in the subgroup, because the inverse of 1 is 1, and the inverse of 4 is 4. Therefore $\{1,4\}$ is a subgroup.

Feedback: Gap/Error. (Associativity is not the same as commutativity.)

$\textbf{EDIT:}$

The definition of associativity is that for all $a,b,c$ in the set, $(ab)c=a(bc)$. But the definition does not say [[a,b,c]] must be distinct. So we can say $\{1,4\}$ is associative because $1(41)=(14)1$,$1(44)=(14)4$,$4(41)=(44)1$,$1(11)=(11)1$, $4(44)=(44)4$, $4(11)=(41)1$, $1(14)=(11)4$, $4(14)=(41)4$.

**Daily Assignment 9/10 $U(7)$**: For the value n=7, determine

-the elements of U(n)

-the order of the group U(n) and the order of each of its elements

-Try to find at least one subgroup

$U(8)=\{1,2,3,4,5,6\}$

1 | 2 | 3 | 4 | ~5 | ~6 | |
---|---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 | 6 |

2 | 2 | 4 | 6 | 1 | 3 | 5 |

3 | 3 | 6 | 2 | 5 | 1 | 4 |

4 | 4 | 1 | 5 | 2 | 6 | 3 |

5 | 5 | 3 | 1 | 6 | 4 | 2 |

6 | 6 | 5 | 4 | 3 | 2 | 1 |

$\mid U(7) \mid = 6$

Order of elements:

$\mid 1 \mid = 1$

$\mid 2 \mid = 3$

$\mid 3 \mid = 6$

$\mid 4 \mid = 3$

$\mid 5 \mid = 6$

$\mid 6 \mid = 2$

One possible subgroup: $\{1,6\}$

In order for $\{1,6\}$ to be a subgroup it must be closed under multiplication, be associative, have an identity, and inverses.

Closure: $\{1,6\}$ is closed under multiplication. When any element of the subgroup is multiplied by itself or another subgroup element, you will always get an element of the subgroup as an answer.

Associativity: $\{1,6\}$ is associative. (1)(6)=(6)(1)

Identity: The identity for $\{1,6\}$ is 1

Inverses: Every element of the subgroup $\{1,6\}$ has an inverse that is an element in the subgroup.

inverse of 1 = 1

inverse of 6 = 6

Therefore, $\{1,6\}$ is one possible subgroup of $U(7)=\{1,2,3,4,5,6\}$

Feedback: Gap/Error. (Associativity is not the same as commutativity.)

$\textbf{EDIT:}$

The definition of associativity is that for all $a,b,c$ in the set, $a(bc)=(ab)c$. But the definition does not say [[a,b,c]] must be distinct. So we can say $\{1,6\}$ is associative because $1(61)=(16)1$,$1(66)=(16)6$,$6(61)=(66)1$,$1(11)=(11)1$, $6(66)=(66)6$, $6(11)=(61)1$, $1(16)=(11)6$, $6(16)=(61)6$.

**Daily Assignment 9/10 $U(9)$**: For the value n=9, determine

-the elements of U(n)

-the order of the group U(n) and the order of each of its elements

-Try to find at least one subgroup

$U(9)=\{1,2,4,5,7,8\}$

1 | 2 | 4 | 5 | 7 | 8 | |
---|---|---|---|---|---|---|

1 | 1 | 2 | 4 | 5 | 7 | 8 |

2 | 2 | 4 | 8 | 1 | 5 | 7 |

4 | 4 | 8 | 7 | 2 | 1 | 5 |

5 | 5 | 1 | 2 | 7 | 8 | 4 |

7 | 7 | 5 | 1 | 8 | 4 | 2 |

8 | 8 | 7 | 5 | 4 | 2 | 1 |

$\mid U(9) \mid = 6$

Order of elements:

$\mid 1 \mid = 1$

$\mid 2 \mid = 6$

$\mid 4 \mid = 3$

$\mid 5 \mid = 6$

$\mid 7 \mid = 3$

$\mid 8 \mid = 2$

One possible subgroup: $\{1,4,7\}$

In order for $\{1,4,7\}$ to be a subgroup it must be closed under multiplication, be associative, have an identity, and inverses.

Closure: $\{1,4,7\}$ is closed under multiplication. When any element of the subgroup is multiplied by itself or another subgroup element, you will always get an element of the subgroup as an answer.

Associativity: $\{1,4,7\}$ is associative. $1 \centerdot (4 \centerdot 7) = (1 \centerdot 4) \centerdot 7$

Identity: The identity for $\{1,4,7\}$ is 1

Inverses: Every element of the subgroup $\{1,4,7\}$ has an inverse that is an element in the subgroup.

inverse of 1 = 1

inverse of 4 = 7

inverse of 7 = 4

Therefore, $\{1,4,7\}$ is one possible subgroup of $U(9)=\{1,2,4,5,7,8\}$

Feedback: Formatting/Writing. Gap/Error.

**3.18**: If $H$ and $K$ are subgroups of $G$, show that $H \bigcap K$ is a subgroup of $G$.

This will be proved by the two step subgroup test. (Theorem 3.2)

First we must show that the subset is nonempty. Since $e$ is an element of both $K$ and $H$, it's also an element of $H \bigcap K$. Therefore the group is nonempty.

Then show that $g$ being an element of $H \bigcap K$ implies that $g^{-1}$ is an element of $H \bigcap K$. Assume $g$ is an element of $H \bigcap K$, then $g$ is an element of $H$. Because $H$ is a subgroup we know all element of $H$ have an inverse element also existing in $H$. Therefore $g^{-1}$ is an element of $H$. Similarly, $g^{-1}$ is an element of $K$. Therefore $g^{-1}$ is an element of $H \bigcap K$.

Then show that $a$ and $b$ being in $H \bigcap K$ implies $ab$ is an element of $H \bigcap K$. Assume $a$ and $b$ are elements of $H \bigcap K$. Then, $a$ and $b$ are elements of $H$. Because $H$ is a subgroup the group has closure, making $ab$ an element of $H$. Similarly, $ab$ is an element of $K$. Therefore $ab$ is an element of $H \bigcap K$.

Therefore $H \bigcap K$ is a subgroup of $G$.

You could then extend this argument by making $H \bigcap K$ intersect another subgroup, which would then again be a subgroup of $G$. Doing this indefinitely will show that any number of intersections will still be a subgroup.

Feedback: $\checkmark$

**3.20**: Let G be a group and let $a \epsilon$ G. Prove that C($a$)=C($a^{-1}$).

We know that the centralizer of a group, or C($a$) = $\{g \epsilon G \mid ga=ag\}$. We must prove that C($a$)=C($a^{-1}$) or in another way of representing it, $\{g \epsilon G \mid ga=ag\}$ = $\{g \epsilon G \mid ga^{-1} =a^{-1}g\}$

Let $g \epsilon$ C($a$). Then $ag=ga$.

Since we know that G is a group, we are able to use the properties of a group: identity, associativity, and inverses on $ag=ga$.

Then,

($a^{-1}$)$ag$=($a^{-1}$)$ga$

$g=(a^{-1})ga$

$g(a^{-1})=a^{-1}ga(a^{-1})$

$ga^{-1}=a^{-1}g$

This shows us that if $g \epsilon$ C($a$) then $ga^{-1}=a^{-1}g$. This is eqiuvalent to $g \epsilon$C($a^{-1}$) since C($a^{-1}$)=$\{g \epsilon G \mid ga^{-1} =a^{-1}g\}$

Now we must show that $g \epsilon$C($a^{-1}$) implies $g \epsilon$C($a$). By steps similar to those above we can see that this is also true.

$ga^{-1}=a^{-1}g$

$ga^{-1}(a)=a^{-1}g(a)$

$g=a^{-1}ga$

$(a)g=(a)a^{-1}ga$

$ag=ga$.

This gives us the double inclusion of the sets, therefore C($a$)=C($a^{-1}$)

Old (before 10/20/2013) Feedback: Gap/Error. (Set equality arguments involve two directions.)

**3.23**: Suppose G is the group defined by the following Cayley table.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
---|---|---|---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

2 | 2 | 1 | 8 | 7 | 6 | 5 | 4 | 3 |

3 | 3 | 4 | 5 | 6 | 7 | 8 | 1 | 2 |

4 | 4 | 3 | 2 | 1 | 8 | 7 | 6 | 5 |

5 | 5 | 6 | 7 | 8 | 1 | 2 | 3 | 4 |

6 | 6 | 5 | 4 | 3 | 2 | 1 | 8 | 7 |

7 | 7 | 8 | 1 | 2 | 3 | 4 | 5 | 6 |

8 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

a. Find the centralizer of each member of G.

b. Find $Z(G)$.

c. Find the order of each element of G.How are these orders arithmetically related to the order of the group?

**a.**

$C(1)=\{1,2,3,4,5,6,7,8\}$

$C(2)=\{1,2,5,6\}$

$C(3)=\{1,3,5,7\}$

$C(4)=\{1,4,5,8\}$

$C(5)=\{1,2,3,4,5,6,7,8\}$

$C(6)=\{1,2,5,6\}$

$C(7)=\{1,3,5,7\}$

$C(8)=\{1,4,5,8\}$

**b.**

$Z(G)=\{1,5\}$

Row 1 is identical to column 1.

Row 5 is identical to column 5.

**c.**

$\mid G \mid = 8$

Order of elements:

$\mid 1 \mid = 1$

$\mid 2 \mid = 2$

$\mid 3 \mid = 4$

$\mid 4 \mid = 2$

$\mid 5 \mid = 2$

$\mid 6 \mid = 2$

$\mid 7 \mid = 4$

$\mid 8 \mid = 2$

The order of the group can be divided by each of the orders of elements and result in one of the orders or itself as a result.

Feedback" $\checkmark$

**3.28**: Must the center of a group be Abelian?

Say $Z(G)$ is not Abelian.

Therefore, there is some $a,x\in Z(G)$ such that $ax\neq xa$.

By Theorem 3.5, because $Z(G)\leq G$, then $a,x\in G$.

By definition of the center of a group, $Z(G)$ consists of all $a\in G$ such that $ax=xa$ for all $x\in G$.

Because of this contradiction, the center of a group must be Abelian.

Feedback: $\checkmark$ (Note that $Z(G) \subseteq G$ by definition.)

**3.51:** Let

under addition. Let

(7)Prove that *H* is a subgroup of *G*. What if 0 is replaced by 1?

To prove that *H* is a subgroup of *G*, we'll use Theorem 3.2, the Two-Step Subgroup Test. This way, we only have to show that *H* is a nonempty subset of *G*, that $AB \in H$ whenever $A, B \in H$, and that $A^{-1} \in H$ whenever $A \in H$.

Note that by the definition of *H*, that each matrix in *H* is also in *G*, we already know that $H \subseteq G$, thus we only need to prove that *H* is nonempty. To show this, we can identify the identity *E* that applies to both sets:

since $0 \in \mathbb{Z}$, and $0 + 0 + 0 + 0 = 0$, and $AE = EA = A$ for any matrix $A \in H$. Thus, *H* is a nonempty subset of *G*.

Let's define two matrices in *H*:

where $a, b, c, d, e, f, g, h \in \mathbb{Z}$, and $a + b + c + d = e + f + g + h = 0$. Thus, we need to show that $AB \in H$ where

(10)We show this by realizing that

(11)Thus, $AB \in H$.

Lastly, for $A \in H$, we must show that $A^{-1} \in H$, which we can use the fact that $AA^{-1} = E$. Thus we have:

(12)To show that $A^{-1} \in H$, we realize that $-a - b - c - d = -1 * (a + b + c + d) = -1 * 0 = 0$. Thus, $A^{-1} \in H$.

Therefore, we have shown that, according to Theorem 3.2, *H* is a subgroup of *G*.

If we were to revise it so that

(13)we wouldn't even be able to find an identity because it requires at least one matrix element in *E* to be non-zero, and since matrix addition is simply addition of its elements, there will be at least one matrix element in the sum *AE* that is not equal to its corresponding matrix element in *A*. So *H* would not be a subgroup of *G*.

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**4.8**: Let $a$ be an element of a group and let $\lvert a\rvert=15$. Compute the orders of the following elements of $G$.

a. $a^3,a^6,a^9,a^{12}$

b. $a^5, a^{10}$

c. $a^2, a^4, a^8, a^{14}$

We use Theorem 4.2, which states that if $a$ is an element of order $n$ in a group and $k$ is a positive integer, then $<a^k>=<a^{gcd(n,k)}>$ and $\lvert a^k \rvert=n/{gcd(n,k)}.$ We are given that the order of $a$ is 15, therefore $n=15$. To find the order of $a^k$, all we need to do is compute $n/gcd(n,k)$, which gives the following solutions.

a. 5

b. 3

c. 15

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**4.9**: How many subgroups does $Z_{20}$ have? List a generator for each of the subgroups. Suppose that $G=<a>$ and $<a>=20$. How many subgroups does $G$ have? List a generator for each of the subgroups.

We use the fundamental theory of cyclic groups and its corollary. According to the corollary, the set $<n/k>$ where $k$ is a positive divisor of $n$ contains the unique and only subgroups of $Z_n$; the subgroups will also have order $k$. We first find all the divisor of 20. They are 1, 20, 2, 10, 5, 4. Therefore, we calculate $n/k$ for each $k$ and we find that the subgroups of $Z_{20}$ are:

$<20>$ of order 1;

$<1>$ of order 20;

$<10>$ of order 2;

$<2>$ of order 10;

$<4>$ of order 5;

$<5>$ of order 4.

For the second part of the problem, we use the Fundamental Theory of Cyclic groups, which states that if the order of a cyclic group $a$ is $n$, then the order of a subgroup is $k$ where $k$ is a divisor of $n$. Also, there is only one subgroup for each $k$. In the problem, $n=20$, so the $k$ values are $1, 20, 2, 10, 4, 5.$ There is exactly one subgroup for each $k$, which makes $6$ different subgroups. The generators of each subgroup will be of the form $a^{n/k}$, which means the generators are:

$a^{20}$

$a^1$

$a^{10}$

$a^2$

$a^5$

$a^4$

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**4.14**: Suppose that a cyclic group $G$ has exactly three subgroups: $G$ itself, ${e}$, and a subgroup of order 7. What is $|G|$? What can you say if 7 is replaced with $p$ where $p$ is a prime?

We know the group $G$ in finite by the first part of Theorem 4.1 and the fact that there is a subgroup of order 7. Since the question implied that $G$ and the subgroup of 7 were distinct groups, we know the order of $G$ is not 7.

Lets say $k$ divides $n$ where $n=|G|$ and $<a>=G$ and $k$ is not equal to 1 or $n$. We know $|<a^{n/k}>|=n/gcd(n,n/k)=n/(n/k)=k$. This implies $k=7$ since $<a^{n/k}>$ will generate a proper subgroup and the only proper subgroup has order 7. Since there are no other nontrivial subgroups of $G$, then there can't be any other divisors of G other than 7.

This leaves only one option for the order of $G$: $|G|=7^{2}=49$.

If you replace 7 with a prime $p$ then $|G|=p^{2}$ via the same argument.

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**4.18**: If a cyclic group has an element of infinite order, how many elements of finite order does it have?

One.

Since there is an element of infinite order (call it $g$), the group it's in has infinite order. This is because the group is cyclic, which means there exists an $a$ so that $<a>$ generates the whole group. This means that for some integer $k$, $a^{k}=g$. Now lets assume $|a|$ is finite and prove it can't be. If $a$ is finite then for some integer $r$, $a^{r}=e$. Then we know $e=(a^{r})^{k}=(a^{k})^{r}=g^{r}$. But this can't be true since the order of $g$ is infinite, therefor $a$ has infinite order. Since we said $a$ generates the group, the group has infinite order.

Since any element can be expressed as a composition of $a$, then no element of the group has finite order. Let's assume element $|b|$ is finite and prove it can't. If $b$ is finite then $b^{x}=e$ and $a^{y}=b$ for integers $x$ and $y$. Then we can show that $(a^{y})^{x}=e$, which contradicts what be know about $|a|$ being infinite. Therefore, any element $b$ in the group must have infinite order.

Except for one, $e$, which by definition will have order 1, giving the result stated above.

Feedback: $\checkmark$ (You can simplify this argument.)

**4.24:** For any element *a* in any group *G*, prove that $<a>$ is a subgroup of *C(a)* (the centralizer of *a*).

Remember that $C(a) = \{ g \in G | ga = ag \}$, or the set of elements from the group *G* where commutativity holds in the operation between each of those elements and *a*. This proof calls on two theorems, and may seem to lack enough explanation. I suggest reviewing the proofs behind the theorems for further clarification.

Theorem 3.6 states that for each element *a* in a group *G*, *C(a)* is a subgroup of *G*. This implies that *C(a)* is a group with the same binary operation of group *G*, as being a group is a property of being a subgroup. We can apply this to Theorem 3.4 (replacing *G* with *C(a)*), stating that because *C(a)* is a group, then if $a \in C(a)$, $<a>$ is a subgroup of *C(a)*. We have not yet explicitly shown that $a \in C(a)$, which is all that is left to show.

Applying the definition of *C(a)*, $a \in G$ and $aa = aa$. The last expression is where we substituted the *g* in the definition for *a*, and may look confusingly simplistic. Another way to explain what it means is that commutativity holds when any element operates on itself. Thus, $a \in C(a)$.

Therefore, according to Theorem 3.4 as we mentioned, $<a>$ is a subgroup of *C(a)*.

Feedback: $\checkmark$

**4.32:** Determine the subgroup lattice for Z_{12}

$\mid Z_{12} \mid$ = 12, note <5> is a generator of $\mid Z_{12} \mid$ and $\mid <5> \mid$ = 12

Divisors of 12 in $\mid Z_{12} \mid$ are 0, 1, 2, 3, 4, 6. We know this from Theorem 4.3.

1: $5^{12/1} = 5^12 = 0, <0>={0, 0, 0, ..., 0}$

2: $5^{12/2} = 5^6 = 6, <6> = {0, 6}$

3: $5^{12/3} = 5^4 = 8, <8>= {0, 8, 4}$

4: $5^{12/4} = 5^3 = 3, <3>= {0, 3, 6, 9}$

6: $5^{12/6} = 5^2 = 10, <10>= {0, 10, 8, 6, 4, 2}$

12: $5^{12/12} = 5^1 = 5, <5>={1, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7}$

<5>

$\diagup$ $\diagdown$

<10> <3>

$\mid$ $\diagdown$ $\mid$

<8> <6>

$\diagdown$ $\diagup$

<0>

This is a subgroup lattice for $\mid Z_{12} \mid$ because <10>, <3>, <8>, <6>, and <0> are all contained in <5>. <8> and <6> are contained in <10>, <6> is contained in <3>, and <0> is contained in <8> and <6>.

**4.38**: Consider the set $\{4, 8, 12, 16\}$. Show that this set is a group under multiplication modulo 20 by constructing its Cayley Table. What is the identity element? Is the group cyclic? If so, find all of its generators.

The Cayley Table to show that this set is a group is shown below:

4 | 8 | 12 | 16 | |
---|---|---|---|---|

4 | 16 | 12 | 8 | 4 |

8 | 12 | 4 | 16 | 8 |

12 | 8 | 16 | 4 | 12 |

16 | 4 | 8 | 12 | 16 |

From this Cayley Table, we can see that the identity element is 16, since the row and column of element 16 in the same order as our original elements. Also, it is true that when 16 is the identity that $a$(16) = 16($a$) = $e$.

To figure out if this set is cyclic, we can find if there is an element $a$ in the set such that $\{4,8,12,16\}$ = $\{a^n | n \in Z\}$. Such an element $a$ is called a generator, which we also must find.

From looking at the powers of each of our $a$ elements, we get:

<4> = $\{16, 4, 16, 4,...\}$

This is not a generator because the subset of numbers does not contain all elements in the original set.

<8> = $\{16, 8, 4, 2, ...\}$

This is a generator because the subset of numbers does contain all elements in the original set. Therefore, from this, we know that our group is cyclic.

<12> = $\{16, 12, 4, 8,...\}$

This is a generator because the subset of numbers does contain all elements in the original set.

<16> = $\{16, 16, 16, ...\}$

This is not a generator because the subset of numbers does not contain all elements in the original set.

From this, we found that our group is cyclic and our generators are 8 and 12.

Feedback: $\checkmark$ (Recheck last line of first paragraph.)

**4.62**: Let $a$ be a group element such that |$a$| = 48. For each part find a divisor $k$ of 28 such that

a.) <$a^{21}$> = <$a^k$>

b.) <$a^{14}$> = <$a^k$>

c.) <$a^{18}$> = <$a^k$>

By Theorem 4.2, we know that <$a^k$> = <$a^{gcd(n,k)}$> when $a$ is an element of order $n$ in a group and $k$ is a positive integer.

From this we can begin with the first part.

In part a, we know from this theorem that to find a divisor $k$ we must solve <$a^k$> = <$a^{gcd(n,k)}$> where our $n$ = 48 since |$a$| = 48. Then, we get <$a^k$> = <$a^{gcd(48,21)}$>. We know that the gcd of 48 and 21 is 3, so we get $k$ = 3.

In part b, we know from this theorem that to find a divisor $k$ we must solve <$a^k$> = <$a^{gcd(n,k)}$> where our $n$ = 48 since |$a$| = 48. Then, we get <$a^k$> = <$a^{gcd(48,14)}$>. We know that the gcd of 48 and 14 is 2, so we get $k$ = 2.

In part c, we know from this theorem that to find a divisor $k$ we must solve <$a^k$> = <$a^{gcd(n,k)}$> where our $n$ = 48 since |$a$| = 48. Then, we get <$a^k$> = <$a^{gcd(48,18)}$>. We know that the gcd of 48 and 18 is 6, so we get $k$ = 6.

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**5.5**: What is the order of the product of a pair of disjoint cycles of lengths 4 & 6?

12, because the order of a product of disjoint cycles is the least common multiple of the lengths of the cycles, and the least common multiple of 4 and 6 is 12.

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**5.6**: Show that A_{8} contains an element of order 15.

An example of an element in A_{8} with order 15 is (123)(45678).

We will have to show that this permutation is both included in A_{8} and has an order of 15.

By Theorem 5.4, this permutation in A_{8} can be expressed as a product of 2-cycles.

Due to the total number of 2-cycles being even, then (123)(45678) is even, thus belonging in A_{8}.

By Theorem 5.3, the order of the permutation is the least common multiple of the lengths of both cycles.

The first cycle has a length of 3, and the second has a length of 5.

(15)Therefore, (123)(45678) has an order of 15.

This ends the proof.

Feedback: $\checkmark$

**5.10**: Show that a function from a finite set $S$ to itself is one-to-one if and only if it is onto. Is this true when $S$ is infinite?

When $S$ is finite, we can assign the cardinality of $S$ as $n$, where $n$ is a positive integer. We must show that one-to-one implies onto and onto implies one-to-one.

Assume $f$ is a function that takes $S$ to itself and is one-to-one. Since $f$ is one-to-one, every element in the domain will be sent to a distinct element in the range. Since there are $n$ element in the domain, $f$ must cover $n$ element in the range. Since there are only $n$ element in the range, all elements are covered. Therefore the function is onto.

Now assume $g$ is a function that takes $S$ to itself and is onto. To make a proof by contradiction, lets assume two elements of the domain go to the same element of the range. That would mean there are $n-2$ element left in the domain to cover the remaining $n-1$ element in the range. Since $n-1 > n-2$, there are not enough element in the domain to cover the remaining elements in the range. Therefore, two elements of the domain can't go to the same element in the range, making the function one-to-one.

Therefore a function from a finite set $S$ to itself is one-to-one if and only if it is onto.

If the set is infinite this isn't necessarily true. A counter example could be the natural numbers to itself (not including $0$). Let $h(x)=x+1$ and take the natural number to the natural numbers. This function is clearly one-to-one since $h(a)=h(b)$ implies $a+1=b+1$ implies $a=b$ where $a$ and $b$ are natural numbers. But this function is not onto because there is no element in the domain that can reach $1$ in the range.

**5.12**:

If $\alpha$ is even, prove that $\alpha^{-1}$ is even. If $\alpha$ is odd, prove that $\alpha^{-1}$ is odd.

Let $\alpha$ = (a b)(c d)

In this situation $\alpha$ is even. The inverse of $\alpha$ = (b a)(d c). $\alpha^{-1}$ is also even.

$\alpha^{-1}$ can also be thought of as a decomposition of $\alpha$. Therefore, according to Theorem 5.5, if $\alpha$ is even, $\alpha^{-1}$ is even.

Let $\alpha$ = (a b)(c d)(e f)

In this situation $\alpha$ is odd. The inverse of $\alpha$ = (b a)(d c)(f e). $\alpha^{-1}$ is also odd.

$\alpha^{-1}$ can also be thought of as a decomposition of $\alpha$. Therefore, according to Theorem 5.5, if $\alpha$ is odd, $\alpha^{-1}$ is odd.

$\textbf{EDIT:}$

Let $\alpha$ = $\beta_1 \beta_2 \cdots \beta_n$ when $n$ is even and each $\beta$ is a 2-cycle. We then know that when we have a 2-cycle, the inverse of the 2-cycle just switches the order of the elements in the permutation. For example, the inverse of (a b) is (b a) and therefore the inverse is still a 2-cycle. So, for each 2-cycle that makes up $\alpha$, each of the inverses are also 2-cycles. Therefore, $\alpha^{-1}$ = $(\beta_1)^{-1} (\beta_2)^{-1} \cdots (\beta_n)^{-1}$ where $n$ is still even and each $\beta$ is a 2-cycle since each inverse is a 2-cycle. Therefore, if $\alpha$ is even, $\alpha^{-1}$ is even.

Let $\alpha$ = $\beta_1 \beta_2 \cdots \beta_n$ when $n$ is odd and each $\beta$ is a 2-cycle. We then know that when we have a 2-cycle, the inverse of the 2-cycle just switches the order of the elements in the permutation. For example, the inverse of (a b) is (b a) and therefore the inverse is still a 2-cycle. So, for each 2-cycle that makes up $\alpha$, each of the inverses are also 2-cycles. Therefore, $\alpha^{-1}$ = $(\beta_1)^{-1} (\beta_2)^{-1} \cdots (\beta_n)^{-1}$ where $n$ is still odd and each $\beta$ is a 2-cycle since each inverse is a 2-cycle. Therefore, if $\alpha$ is odd, $\alpha^{-1}$ is odd.

Feedback: Gap/Error. (Generalize!)

**5.17**:

Let

and

Compute the following:

a. $\alpha^{-1}$

b. $\beta\alpha$

c. $\alpha\beta$

Write $\alpha$ and $\beta$ in cycle notation.

a.

(18)$\alpha = (1,2)(4,5)$

$\beta = (1,6,5,3,2)$

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**5.20**: Compute the order of each member of $A$_{4}. What arithmetic relationship do these orders have with the order of $A$_{4}?

According to Theorem 5.3 and reference to Table 5.1, the orders of the elements of $A$_{4} are as follows:

|$\alpha$_{1}|=1

|$\alpha$_{2}|=|$\alpha$_{3}|=|$\alpha$_{4}|=2

|$\alpha$_{5}|=|$\alpha$_{6}|=|$\alpha$_{7}|=|$\alpha$_{8}|=|$\alpha$_{9}|=|$\alpha$_{10}|=|$\alpha$_{11}|=|$\alpha$_{12}|=3

By Theorem 5.7, the order of $A$_{4} is 12.

The orders of the elements are all divisors of the order of the group.

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**5.22**: Let α and β belong to S_{n}. Prove that α^{-1}β^{-1}αβ is an even permutation.

Since a Permutation such as these can be written as a combination of 2-cycles, and can be even or odd depending on how many there are, we need to show that this combination of 4 permutations is even. Since an even integer is represented by 2k, where k can be odd or even also. Since the inverses of α and β will be the same even or oddness as α and β, we can generalize by letting a be the number of 2-cycles represented by α and b be the number of 2-cycles of β, so we have abab, or if these are integers, 2a2b. 4(ab), and since 4 times any integer is even, α^{-1}β^{-1}αβ is an even permutation.

**5.24**: How many elements of order 5 are in S_{7}?

Elements in S_{7} with order 5 have to be 5-cycles, with 2 other 1 cycles, because we cannot any other combination of cycles because for their order to equal 5, their lengths would have to equal 5 when multiplied and since 5 is prime we have no such combinations. So we need to know how many different 5-cycles can be made. Well there are 7*6*5*4*3 ways to make 5 cycles from S_{7}, but we cannot count the same cycles twice, since (1 2 3 4 5) = (2 3 4 5 1), so since there are 5 ways to write the same cycle, we just divide 7*6*5*4*3 by 5, so there are 504 elements of order 5 in S_{7}.

**5.37**: Suppose that B is a 10-cycle. For which integers i between 2 and 10 is B^{i} also a 10-cycle?

For B^{i} to also be a 10-cycle, the integer i will have to be relatively prime to 10, because B has order 10, so B^{i} has order 10/GCD(10,i) according to theorem 4.2. So we need the integers i's so that GCD(10,i) is 1, which are 3,7, and 9, the relatively prime integers that have a GCD of 1 with 10.

**5.40**: Represent the symmetry group of an equilateral triangle as a group of permutations of its vertices (See Exercise 3).

From looking at the symmetries of an equilateral triangle with vertices labeled 1, 2, and 3 with 1 being the topmost vertex, 2 being the bottom left vertex, and 3 being the bottom right vertex, we are able to represent the symmetry group of this triangle as a group of permutations of its vertices. We know the 6 reflections and rotations of an equilateral triangle are Id (the identity), $R_{120}$ (rotation of 120$^{\circ}$), $R_{240}$ (rotation of 240$^{\circ}$), V (vertical reflection), D (a diagonal reflection), and D' (another diagonal reflection).

Beginning with the identity, as described above, 1 goes to 1, 2 goes to 2, and 3 goes to 3 yielding (1)(2)(3) as the permutation or

(21)Next we must look at $R_{120}$. If we were to draw this triangle with the numbers, 3 would be at the top, 1 on the bottom left, and 2 at the bottom right. Looking from our new reflection back to our identity, 1 appears in the 2 spot on the identity, 2 appears in the 3 spot on the identity, and 3 appears in the 1 spot on the identity. In other words, 1 goes to 2, 2 goes to 3, and 3 goes to 1 yielding (123) or

(22)Next we must look at $R_{240}$. If we were to draw this triangle with the numbers, 2 would be at the top, 3 on the bottom left, and 1 at the bottom right. Looking from our new reflection back to our identity, 1 appears in the 3 spot on the identity, 2 appears in the 1 spot on the identity, and 3 appears in the 2 spot on the identity. In other words, 1 goes to 3, 2 goes to 1, and 3 goes to 2 yielding (132) or

(23)Next we must look at V. If we were to draw this triangle with the numbers, 1 would be at the top, 3 on the bottom left, and 2 at the bottom right. Looking from our new reflection back to our identity, 1 appears in the 1 spot on the identity, 2 appears in the 3 spot on the identity, and 3 appears in the 2 spot on the identity. In other words, 1 goes to 1, 2 goes to 3, and 3 goes to 2 yielding (1)(23) or

(24)Next we must look at D. If we were to draw this triangle with the numbers, 2 would be at the top, 1 on the bottom left, and 3 at the bottom right. Looking from our new reflection back to our identity, 1 appears in the 2 spot on the identity, 2 appears in the 1 spot on the identity, and 3 appears in the 3 spot on the identity. In other words, 1 goes to 2, 2 goes to 1, and 3 goes to yielding (12)(3) or

(25)Next we must look at D'. If we were to draw this triangle with the numbers, 3 would be at the top, 2 on the bottom left, and 1 at the bottom right. Looking from our new reflection back to our identity, 1 appears in the 3 spot on the identity, 2 appears in the 2 spot on the identity, and 3 appears in the 1 spot on the identity. In other words, 1 goes to 3, 2 goes to 2, and 3 goes to 1 yielding (2)(13) or

(26)Feedback: $\checkmark$

**5.44**: Find a cycle subgroup of $A_8$ that has order 4

We must look at permutations in $S_8$ that has order 4.

We can look at (1234) in $S_8$. This will be our generator. $(1234)^4 = e$. Then, we have (1234) being of order 4 since we must permute (1234) 4 times to get back to the identity. Having this generator with an order of 4 to get back to the identity makes this cyclic.

We can create a subgroup of (1234) with 4 elements based off of this original permutation. This subgroup is $\{e, (1234), (1234)^2, (1234)^3\}$.

I claim this is a cyclic subgroup. To prove this, we can use the Finite Subgroup Test to prove closure in our subgroup and also prove our subgroup is nonempty.

We have 4 elements in our subgroup, $e, (1234), (1234)^2, (1234)^3$ so this proves our subgroup is nonempty.

Now, we must show we have closure in our subgroup by showing each element being permuted with each other element.

We can first rewrite each of our elements as:

$e$=(1)(2)(3)(4)

(1234)=(1234)

$(1234)^2$ = (13)(24)

$(1234)^3$ = (1432)

Then, we can prove closure:

$e$=(1234)=(1)(2)(3)(4)(1234) = (1234)

$e$=$(1234)^2$=(1)(2)(3)(4)(13)(24)=(13)(24)= $(1234)^2$

$e$=$(1234)^3$=(1)(2)(3)(4)(1432)=(1432)= $(1234)^3$

$(1234)(1234)^2$=(1234)(13)(24)=(1432) =$(1234)^3$

$(1234)^2(1234)$=(13)(24)(1234)=(1432)

$(1234)(1234)^3$=(1234)(1432)=(1)(2)(3)(4)=$e$

$(1234)^3(1234)$=(1432)(1234)=(1)(2)(3)(4)=$e$

$(1234)^2(1234)^3$=(13)(24)(1432)=(1234)

$(1234)^3(1234)^2$=(1432)(13)(24)=(1234)

From all of this, we prove that the subgroup $\{e, (1234), (1234)^2, (1234)^3\}$ is cyclic and in $A_8$

**5.45**: Find a noncyclic subgroup of $A_8$ that has order 4.

$(1234)(56)$ is a noncyclic subgroup of order 4. Because $A_8$ is the set of all even permutations of $S_n$ and $(1234)(56)$ can be written as $(12)(23)(34)(56)$ which is an even number of 2-cycles and thus an even permutation, $(1234)(56)$ is a nonempty finite subset of $A_8$. Because $(1234)(56)$ is closed under the operation of permutations (because taking permutations of cycles will not introduce new elements), we know that $(1234)(56)$ is a subgroup of $A_8$ by the finite subgroup test. We also know that the order is 4 by Theorem 5.3, because the LCM of the cycle lengths 2 and 4 (which we get when the subgroup is written in disjoint cycle notation) is 4. Finally, we know that the subgroup is noncyclic because there is no element $a$ such that $a^n$ for integer $n$ generates the entire subgroup—for instance, a member of the $(56)$ cycle can only generate itself and the other member of the $(56)$ cycle, and any member of the $(1234)$ cycle can only generate itself and the other members of the $(1234)$ cycle.

Feedback: Gap/Error. (Be careful to distinguish between an element and a subgroup.)

**EDIT:**

The subgroup

$H=\{(1),(123),(132),(456),(465),(123)(456),(132)(456),(123)(465),(132)(465)\}$

is a noncyclic subgroup of $A$_{$8$}.

$(1)$ serves as the identity.

Each element has an inverse:

$a$_{$1$}$=e$

$a$_{$2$}$a$_{$3$}$=e$

$a$_{$4$}$a$_{$5$}$=e$

$a$_{$6$}$a$_{$9$}$=e$

$a$_{$7$}$a$_{$8$}$=e$

for each corresponding element in $H$.

Since every element in $H$ has the same order (except for the identity), $H$ is associative.

Every element is either a 3-cycle or a product of 3-cycles. Since a 3-cycle can be written as a product of 2 nondisjoint 2-cycles, that means every 3-cycle or product of 3-cycles is even, thus belonging in $A$_{$8$}.

Finally, $H$ is noncyclic because there is no element in $a\in H$ such that $a$^{$n$} equals every element in $H$ for any $n\in Z$.

**5.47**: Show that every element in $A$_{$n$} for $n \geq 3$ can be expressed as a 3-cycle or a product of three cycles.

Every permutation in $A$_{$n$} that is either a 3-cycle or a product of only three cycles is already proven.

Now we must show that every other permutation of $A$_{$n$} can be written as a 3-cycle or a product of 3-cycles.

Because the order of a 3-cycle is 3, we can say that

$e = (a$_{$1$},$a$_{$2$},$a$_{$3$}$)(a$_{$1$},$a$_{$2$},$a$_{$3$}$)(a$_{$1$},$a$_{$2$},$a$_{$3$}$)$

and that takes care of the identity.

By Theorem 5.4, every permutation in $A$_{$n$} can be written as a product of 2-cycles. Because it is $A$_{$n$}, every permutation is even and therefore has an even number of 2-cycles.

We will prove the rest by induction. Let's assume that there is an element in $A$_{$n$} made up of exactly 2 2-cycles.

We can assume that this pair of 2-cycles is either disjoint or not disjoint.

Say the pair is not disjoint. It can look either one of three ways.

If the pair looks like this

$(a$_{$1$},$a$_{$2$}$)(a$_{$1$},$a$_{$2$}$)$

then the permutation is the identity and we have already proven for that.

If the pair looks like this

$(a$_{$1$},$a$_{$2$}$)(a$_{$1$},$a$_{$3$}$)$

or this

$(a$_{$1$},$a$_{$2$}$)(a$_{$2$},$a$_{$3$}$)$

then it can be written as a 3-cycle as such

$(a$_{$1$},$a$_{$3$},$a$_{$2$}$)$

or as such

$(a$_{$1$},$a$_{$2$},$a$_{$3$}$)$

respectively, proving that two non-disjoint cycles can be written as a 3-cycle.

Say the pair is disjoint. Therefore, the permutation will look like this:

$(a$_{$1$},$a$_{$2$}$)(a$_{$3$},$a$_{$4$}$)$.

We can make this the same as a product of 3-cycles, as long as 1 goes to 2 and 3 goes to 4 in the end.

For example:

$(a$_{$2$},$a$_{$4$},$a$_{$3$}$)(a$_{$1$},$a$_{$3$},$a$_{$2$}$)$.

In here, 1 goes to 3, which goes to 2. Looking for 2, we see it only goes to 1, and it is fixed in the left-most cycle.

And then, 3 goes to 2, and 2 goes to 4. Looking for 4, we see it is fixed in the right cycle, and then goes to 3.

We have replicated the 2-cycle product of $(a$_{$1$},$a$_{$2$}$)(a$_{$3$},$a$_{$4$}$)$. This ends the base case of the induction proof.

For the next step of the induction, we look at a permutation with exactly 4 2-cycles, since we are dealing with only even permutations.

We can use the base case on the first two 2-cycles. It will either result in 1 3-cycle or 2 3-cycles. As long as the result of 3-cycles are equal to the 2-cycles, then the new result combined with the second two 2-cycles will yield the same permutation. Therefore, we can use the base case on the second two 2-cycles and come up with a combination of two, three, or four 3-cycles.

By induction, this proves that every permutation in $A$_{$n$} can be written as a 3-cycle or as a product of 3-cycles.

**5.48:** Show that for $n \geq 3$, $Z(S_n) = \{\varepsilon\}$.

Recall that our definition for the center of a group *G* is $Z(G) = \{a \in G | ax = xa$ for all $x \in G\}$. For this situation, I use an interesting proof by induction due to the fact that the induction case doesn't obviously need the statement to be true for *n* for it to be true for *n* + 1, although convention and technicalities certainly call for us to take this traditional route.

For both the base case and the induction case, we simply need need to explain why $\varepsilon$ fits the definition of $Z(S_n)$, and show for every other single-cycle permutation element, *a*, an element, *x*, that will not fit the description of $Z(S_n)$. I will explain here once and for all why $\varepsilon \in Z(S_n)$ for $n \geq 2$. The identity permutation $\varepsilon$ is defined such that $\varepsilon \alpha = \alpha \varepsilon = \alpha$ for any permutation $\alpha$. This is because $\varepsilon$ sends every element to itself and thus does not change the overall result. Now we are ready to disprove the other permutations in an induction proof.

**Base Case, n = 3:** We can state that $S_3 = \{\varepsilon, (12), (23), (13), (123), (132)\}$. For our non-identity permutations, we can calculate that:

Remember that for each candidate element for $Z(S_3)$, *a*, we only need to show one element in $G$, *x*, for which the operation is noncommutative. Thus, $Z(S_3) = \{\varepsilon\}$.

**Inductive Case, $Z(S_n) = \{\varepsilon\} \Rightarrow Z(S_{n+1}) = \{\varepsilon\}$:** We must split this generalized case into 2-cycles and *k*-cycles for $2 < k \leq n + 1$.

**2-cycles:** As $n + 1 \geq 4$, we know that for any 2-cycle, there are at least 2 symbols that aren't in that 2-cycle. Thus we can generalize that for symbols *a*, *b*, and *c* that gets permuted by the elements of $S_{n + 1}$:

This implies that no 2-cycles of $S_{n + 1}$ will be elements of $Z(S_{n + 1})$ for $n + 1 \geq 4$.

** k-cycles:** For

*k*-cycles, where $2 < k \leq n + 1$, we calculate that for symbols $a_i$ that gets permuted by the elements of $S_{n + 1}$:

This implies that there are no cycles in $S_{n + 1}$ of length 3 through $n + 1$ that will be an element of $Z(S_{n + 1})$ for $n + 1 \geq 4$. These statements about cycles of length 1 ($\varepsilon$) through $n + 1$ prove to be true for $S_{n + 1}$ given the fact that the similar set of statements are true when we replace $n + 1$ with $n$.

This argument can be extended for multi-cycle permutation elements in $S_n$ for $n \geq 4$. After we convince ourselves of this fact, we see how this shows that for $n \geq 3$, $Z(S_n) = \{\varepsilon\}$.

**5.56**: Given that $\beta$ and $\gamma$ are in $S_4$ with $\beta \gamma = (1432)$ and $\gamma \beta=(1243)$ and $\beta(1)=4$, determine $\beta$ and $\gamma$.

and

[[math]]

\gamma=

\begin{bmatrix}

1 & 2 & 3 & 4\4 & 2 & 1 & 3

\end{bmatrix}

[[/math]].

We find this by starting with the $\beta\gamma$ product. We know that $\beta(1)=4$, and we know that $\beta\gamma(3)=4$, and therefore $\gamma(3)$ must be $1$. Turning to the $\gamma \beta$ product, we know that $\beta(1)=4$ and $\gamma \beta(1)=3$, and therefore $\gamma(4)$ must be $3$. With this knowledge, we go back to the $\beta \gamma$ product and fill in $\gamma(4)=3$. We continue in this manner until we know all the values, which gives the $\beta$ and $\gamma$ values listed above. (We can then multiply our results back out to make sure they give the correct products, which they do.)

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