## 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

**6.01**: Find an isomorphism from the group of integers under addition to the group of even integers under addition.

By looking at the isomorphism from the group (1,2,3,4) to the group (2,4,6,8), we can prove that this is an isomorphism by doing the four steps of showing mapping, one-to-one, onto, and operation-preserving.

When thinking about (1,2,3,4) being mapped to (2,4,6,8), we can find a function to map this, $\phi$(x)=2x. From this, we see that we have a function so this proves the mapping step.

Now we can move on to the step of showing that this function is one-to-one. To do this, we must assume$\phi$(x)=$\phi$(y) and use this to prove that x=y. We start by supposing that 2x=2y. Then, we can divide by 2 on both sides to yield x=y. This proves the function is one-to-one.

Now, we must prove the function is onto. To do this, we must find that for any positive even integer y, there is some positive integer x such that $\phi$(x)=y. We know this is the case when 2x=y, so we have proved the function is onto. EDIT: The function is onto because for any element under the group of even integers, we divide it by half to show the element under the group of integers that gets sent to it.

Lastly, we must prove that the function is operation-preserving. To do this, we must show that $\phi$(xy)=$\phi$(x)$\phi$(y). In this case, we must show, $\phi$(x+y)=$\phi$(x)+$\phi$(y). So $\phi$(x+y)=2(x+y)=2x+2y=$\phi$(x)+$\phi$(y). Therefore, the function is operation-processing.

Since we have shown (1,2,3,4) to (2,4,6,8) has a mapping, is one-to-one, is onto, and is operation-preserving, we have proved it is an isomorphism. EDIT: Note that this argument works for said isomorphism sending the group of all integers to the group of all even integers, and that zero is an even number.

Feedback: Confusing ambiguous. (Your description of the group is incomplete. Review onto argument.)

**Daily 10/2**: Suppose $G$ is a cyclic group of order $n$. Prove that $G$ is isomorphic to $Z_n$.

Since $G$ is a cyclic group, there has to exist a generator element of $G$, lets call is $x$, so that $<x>=G$ and $|x|=n$. Then we can set up a function from $Z_n$ to $G$, $\varphi$. We'll defined $\varphi$ as $\varphi (k)=x^{k}$ where $k$ is an element of $Z_n$. We then must show this function is one-to-one, onto, and operation preserving.

First one-to-one. Assume two $\varphi (a)= \varphi (b)$, where $a$ and $b$ are elements of $Z_n$, and show $a=b$. $\varphi (a)= \varphi (b)$ means $x^{a}=x^{b}$ from how we defined our function, which is true is and only if $n$ divides $a-b$. Since $a$ and $b$ are elements of $Z_n$, without loss of generality, $0 \leq b \leq a \leq n-1$. This makes the difference of $a$ and $b$ no greater then $n-1$, which means the only way $n$ will divide $a-b$ is if $a-b=0$, which implies $a=b$. This proves $\varphi$ is one-to-one.

Then prove onto. We must show for every element $g$ of $G$, there is an element $z$ of $Z_n$, so that $\varphi(z)=g$. Because we said there was an generator, $x$, of $G$, every element of $G$ can be expressed as $x^{m}$ where $m$ is a integer between $0$ and $n-1$. Restated, we're saying $g=x^{m}$ where $g$ is a arbitrary element of $G$. From the definition of $\varphi$, $\varphi(z)=x^{k}$ where $k$ is an element of $Z_n$. Since $Z_n$ has all integers between $0$ and $n-1$, we know we can always express some element of $G$ in the form $\varphi(z)$. Therefore, $\varphi$ is onto.

Now to show $\varphi$ is operation preserving. For any two elements, $c$ and $d$ of $Z_n$, we want to show $\varphi(cd)=\varphi(c) \varphi(d)$. Going off our definition of $\varphi$, we know $\varphi(c) \varphi(d)=x^{c}x^{d}=x^{c+d, mod(n)}$. We also know $cd=c+d, mod(n)$ which then implies $\varphi(cd)=x^{c+d, mod(n)}$. Therefore $\varphi$ is operation preserving.

Therefore, $G$ is isomorphic to $Z_n$.

Feedback: $\checkmark$

**6.4**: Show that $U(8)$ is not isomorphic to $U(10)$

$U(8)$ | 1 | 3 | 5 | 7 |
---|---|---|---|---|

1 | 1 | 3 | 5 | 7 |

3 | 3 | 1 | 7 | 5 |

5 | 5 | 7 | 1 | 3 |

7 | 7 | 5 | 3 | 1 |

$U(10)$ | 1 | 3 | 7 | 9 |
---|---|---|---|---|

1 | 1 | 3 | 7 | 9 |

3 | 3 | 9 | 1 | 7 |

7 | 7 | 1 | 9 | 3 |

9 | 9 | 7 | 3 | 1 |

$\phi$ 1 = 1

$\phi$ 9 = 1

In order for $U(8)$ to be isomorphic to $U(10)$, $\phi$ must be one-to-one. That is, all distinct elements must go to distinct elements. Since $\phi$ 1 = $\phi$ 9 = 1 and 1 $\neq$ 9, $\phi$ is not one-to-one and $U(8)$ is not isomorphic to $U(10)$

**EDIT:** To prove that $U(8)$ is not isomorphic to $U(10)$, it is enough to show that the two Groups do not have an equal number of elements of equal order, by theorem 6.2 number 7.

The order of elements in $U(8)$ are:

|1| = 1

|3| = 2

|5| = 2

|7| = 2

While the order of elements in $U(10)$ are:

|1| = 1

|3| = 4

|7| = 4

|9| = 2

Therefore, no $\phi$ exists

Feedback:Gap/Error. (Are you claiming that $\phi(9)=1$ for every possible map between the groups?)

**6.05**: Show that $U(8)$ is isomorphic to $U(12)$.

$U(8)=\{1,3,5,7\}$ and $U(12)=\{1,3,7,11\}$. We try to construct a mapping $\phi$ from $U(8)$ to $U(12)$. We define:

$\phi(1)=1$

$\phi(3)=5$

$\phi(5)=7$

$\phi(7)=11$

Because no two elements map to the same element, $\phi$ is one-to-one. Because at least one element of $U(8)$ maps to each element of $U(12)$, $\phi$ is onto. Therefore, $\phi$ is bijective.

Lastly, we must show that $\phi$ is operation preserving: that is, that $\phi(gh)=\phi(g) \phi(h)$ for all $g,h$ in $U(8)$.

We calculate all possible pairs of $g$ and $h$.

$\phi((1)(3))=\phi(3)=5$, which is the same as $\phi(1)\phi(3)=(1)(5)=5$;

$\phi((1)(5))=\phi(5)=7$, which is the same as $\phi(1)\phi(5)=(1)(7)=7$;

$\phi((1)(7))=\phi(7)=11$, which is the same as $\phi(1)\phi(7)=(1)(11)=11$;

$\phi((3)(5))=\phi(7)=11$, which is the same as $\phi(3)\phi(5)=(5)(7)=11$;

$\phi((3)(7))=\phi(5)=7$, which is the same as $\phi(3)\phi(7)=(5)(11)=7$;

$\phi((5)(7))=\phi(3)=5$, which is the same as $\phi(5)\phi(7)=(7)(11)=5$;

$\phi((1)(1))=\phi(1)=1$, which is the same as $\phi(1)\phi(1)=(1)(1)=1$;

$\phi((3)(3))=\phi(1)=1$, which is the same as $\phi(3)\phi(3)=(5)(5)=1$;

$\phi((5)(5))=\phi(1)=1$, which is the same as $\phi(5)\phi(5)=(7)(7)=1$;

$\phi((7)(7))=\phi(1)=1$, which is the same as $\phi(7)\phi(7)=(11)(11)=1$.

Because $\phi$ is operation preserving for all pairs of $g,h$ in $U(8)$, $\phi$ is operation preserving for the mapping, and therefore $\phi$ is an isomorphism.

Feedback: $\checkmark$

**6.9:** In the notation of Theorem 6.1, prove that $T_e$ is the identity and that $(T_g)^{-1} = T_{g^{-1}}$.

Given Theorem 6.1, Cayley's Theorem, we know that $\bar{G} = \{T_g | g \in G\}$, where $T_g (x) = gx$ for $x \in G$.

**Identity:** With this and the rest of the theorem, we know that $T_e (x) = ex = x$, and that:

for all $x \in G$. Because each range element is mapped by one given domain element for all three of the functions, which is how we define function composition, we can say that:

(2)where the first $e$ is the identity element of $G$, and the second $e$ represents the identity element of $\bar{G}$, and thus so does $T_e$.

**Inverse:** Due to the theorem and the identity of $\bar{G}$, we know that $T_g (T_g)^{-1} = T_e$. Using this, we show that:

for all $x \in G$. Again, because of how we define function composition, we find that:

(4)Note that we never had to discuss isomorphisms, which is the found throughout Cayley's Theorem.

Feedback: Confusing/ambiguous. (Carefully think through what it means to be the identity, an inverse in $\overline{G}$.)

**6.11**: For inner automorphisms $\phi$_{$g$}, $\phi$_{$h$}, and $\phi$_{$gh$}, prove that $\phi$_{$g$}$\phi$_{$h$}=$\phi$_{$gh$}.

$\phi$_{$g$}$(x) = gxg$^{$-1$}, $\phi$_{$h$}$(x) = hxh$^{$-1$} and $\phi$_{$gh$}$(x) = (gh)x(gh)$^{$-1$}.

Because $g$ and $h$ are elements of a group, $(gh)$^{$-1$}$= g$^{$-1$}$h$^{$-1$}.

$\phi$_{$g$}$\phi$_{$h$}$(x)$ = $\phi$_{$g$}$(\phi$_{$h$}$(x))$ = $\phi$_{$g$}$(hxh$^{$-1$}$)$ = $ghxh$^{$-1$}$g$^{$-1$} = $(gh)x(gh)$^{$-1$} = $\phi$_{$gh$}$(x)$.

Therefore, $\phi$_{$g$}$\phi$_{$h$}=$\phi$_{$gh$}.

Feedback: $\checkmark$

**6.14:** Find $Aut(\mathbb{Z}_6)$

In order to make $\alpha$ an automorphism of $\mathbb{Z}_6$ we must determine the choices for $\alpha(1)$. Property 5 of Theorem 6.2 tells us that $\mid \alpha(1) \mid$ = 6. There are two candidates for $\alpha(1)$: $\alpha(1) = 1 = \alpha_1$ and $\alpha(1) = 5 = \alpha_5$. $\alpha_1$ is the identity. Since $x$mod6 = $y$mod6 implies 5$x$mod6 = 5$y$mod6, $\alpha_5$ is well defined.

Since $\alpha_5$(1)=5 is a generator of $Aut(\mathbb{Z}_6)$, $\alpha_5$ is onto and one-to-one.

Since $\alpha_5$$(a+b)$ = $5(a+b)$ = $5a+5b$ = $\alpha_5$(a) + $\alpha_5$(b), $\alpha_5$ is operation preserving.

Therefore $\alpha_5$ $\in$ $Aut(\mathbb{Z}_6)$.

$Aut(\mathbb{Z}_6)$ = $\alpha_5$

**6.20**: Suppose that $\phi$: Z_{50} -> Z_{50} is an automorphism with $\phi(11)$ = 13. Determine a formula for $\phi(x)$

We know that 11 is relatively prime to 50, therefore it is a generator of Z_{50}, and we only know that $\phi(11)$ = 13, which we must use to construct a function. For all the elements $x \in Z,,50,,\}$ x = y*11(mod50), and $\phi(x)$=$\phi(y*11)$ = y*$\phi(11)$ = y*13(mod50). We need to know what $\phi(1)$ maps to in order to help us form the entire equation, , so we let x = 1 and solve for y, so 1 = 41*11(mod50), $\phi(41*11)$ = 41*$\phi(11)$ = 41*13 = 33 Therefore, $\phi(x)$ = x*33(mod50) for all $x \in Z,,50,,\}$

**6.21:** Prove property 1 of Theorem 6.3.

Theorem 6.3, property 1, states that if $\phi$ is an isomorphism from a group $G$ onto a group $\bar{G}$, then $\phi^{-1}$ is an isomorphism from $\bar{G}$ onto $G$. To show that the inverse of $\phi$ is such an isomorphism, we need to show how it maps, that it's one-to-one and onto, and that it's operation preserving.

Note that throughout this proof, $\bar{a}, \bar{b} \in \bar{G}$, and $a, b, \in G$.

**Mapping:** We can simply define the mapping of $\phi^{-1}: \bar{G} \rightarrow G$ to be such that $\phi^{-1} (\bar{a}) = a$ for $\bar{a} \in \bar{G}$ and $a \in G$.

**One-to-one:** For this, we must show that $\phi^{-1} (\bar{a}) = \phi^{-1} (\bar{b})$ implies that $\bar{a} = \bar{b}$. Due to our definitions, we can directly rewrite this as $a = b$ implying that $\phi (a) = \phi (b)$, which is certainly true. Thus, so must be the former claim which means that $\phi^{-1}$ is one-to-one.

**Onto:** For this, we must show that for any $a$, $\phi^{-1} (\bar{a}) = a$. Because $\phi$ maps from $G$ onto $\bar{G}$, we know that $\phi (a)$ must be an element in $\bar{G}$. For this element we assign the value $\phi (a) = \bar{a}$. Due to our definition of the inverse function, we see that $\phi^{-1} (\bar{a}) = a$.

**Operation preserving:** We must show that $\phi^{-1} (\bar{a} \bar{b}) = \phi^{-1} (\bar{a}) \phi^{-1} (\bar{b})$. Because we already know that $\phi$ is an isomorphism, we can refer to the operation preserving condition of that function. We find that $\phi(ab) = \phi(a) \phi(b) = \bar{a} \bar{b}$. Using this, we see that:

Feedback: $\checkmark$

**Daily Assignment 10/7**: Write up a careful proof that $\phi(x)=e^x$ is an isomorphism from $R$ to $R^+$

To show $\phi(x)=e^x$ is an isomorphism we must show that it is onto, one to one, and operation preserving.

We will note that $R^+$ is under multiplication and $R$ is under addition.

To show operation preservation we want to show that $\phi(x)\phi(y)=\phi(x+y)$ so we know that $\phi(x)\phi(y)=e^xe^y=e^{x+y}$ and we know that $\phi(x+y)=e^{x+y}$ because $R^+$ are under multiplication and $R$ are under addition. So we have proved that $\phi(x)\phi(y)=\phi(x+y)$.

We can show one-to-one by showing that every element in $R$ goes to a distinct element in $R^+$. So we can assume that $e^x=e^y$ then apply natural log to both sides to get $ln(e^x)=ln(e^y)$ so we know that x=y, which means that x and y must be distinct elements in $R$ and $R^+$.

To show onto we want to show that for every element in $R^+$ there is some element in $R$ thats sent to it. We can do this by letting $\phi(x)=y$ where $x \in R$ and $y \in R^+$, so we know that $y=e^x$, and if we apply the natural log to both sides we get $ln(y) = ln(e^x)$ so $x=ln(y)$ which shows that for every element in $R^+$ there is some element in $R$ that is sent to it. EDIT: Note that an exponent can be positive, negative or zero, and will always give us a positive result. So for every $\overline{a} \in R^+$, we have $a \in R$ such that $\phi(a) = e^a = e^{\ln(\overline{a})} = \overline{a}$. This fulfills the definition of an onto function.

So we have shown that $\phi(x)=e^x$ is an isomorphism from $R$ to $R^+$ because it is a bijection and operation preserving.

Feedback: Confusing/ambiguous. (Are you defining $\overline{a}=\phi(a)$?)

Feedback: Confusing/ambiguous. (Review onto argument.)

**7.2:** Let $H$ be as in Exercise 1. How many left cosets of $H$ in $S$_{$4$} are there? (Determine this without listing them.)

$H=\{(1),(12)(34),(13)(24),(14)(23)\}$

According to Lagrange's Theorem, the number of distinct left cosets of $H$ in $S$_{$4$} is 24/4.

There are 6 left cosets of $H$.

Feedback: $\checkmark$

**7.8:** Suppose $a$ has order 15. Find all of the left cosets of $<a^5>$ in $<a>$.

Because $|<a^5>| = 3$, by Lagrange's Theorem, we know that there will be 5 distinct left cosets of $<a^5>$ in $<a>$. But to find the 4 other than $<a^5>$, we just have to operate 4 elements on $<a^5>$ that are all disjoint from each other.

After some trial and error, we see that the distinct cosets are:

(6)Basically, we can choose to operate the positive powers of $a$ that is less than 5.

**7.10**: Let $a$ and $b$ be nonidentity elements of different orders in a group $G$ of order 155. Prove that the only subgroup of $G$ that contains $a$ and $b$ is $G$ itself.

Let H be a subgroup of G such that $a$, $b$ are in H.

Lagrange's Theorem says that if G is a finite group and H is a subgroup of G, the |H| divides |G| or |G| / |H|. From this, we know that |H| divides |G| and since the order of an element divides the order of the group, |a| and |b| divides |H| which divides |G|. So, $a$ and $b$ must divide 155.

If $a$ or $b$ has order 155, then |H| / 155 so |H| must be 155 since |H| $\leq$ |G| = 155. Therefore, |H| =155 and H=G.

If $a$ = 31, $b$ = 5 or $a$ = 5 , $b$ = 31, then |H| / 5 and |H| / 31. For both of these to be true, |H| = gcd(5,31) =155. Therefore, |H| = 155 and H=G, which is true in both cases.

Therefore, the only subgroup of G that contains $a$ and $b$ is G itself

**7.14**: Suppose that $K$ is a proper subgroup of $H$ and $H$ is a proper subgroup of $G$. If $|K|$ = 42 and $|G|$ = 420, what are possible orders of $H$?

Lagrange's Theorem says that if G is a finite group and H is a subgroup of G, the |H| divides |G| or |G| / |H|.

From, this we know that |K| divides |H| and |H| divides |G|. So, 42 divides |H| and |H| divides 420. It is given that K is a proper subgroup of H, so |H| $\neq$ 42 and it is also given that H is a proper subgroup of G, so |H| $\neq$ 420. To find the possible orders of H, we can look at the multiples of 42 that are less than 420 and divide 420 evenly. From this, we find |H| = 84 or |H| = 210.

**7.15** Let G be a group with $|G|=pq$, where p and q are prime. Prove that every proper subgroup of G is cyclic.

According to Lagrange's Theorem, if G is a finite group and H is a subgroup of G, then $|H|$ divides $|G|$. If $|G|=pq$, then any subgroup of G has an order that must divide $pq$. Since $p$ and $q$ are both prime, the only values that divide $pq$ are $p$, $q$, and 1.

According to Corollary 3, a group of prime order is cyclic. Since the subgroups of $pq$ are $p$, $q$, and 1 and all three are prime values, it can be said that all three are cyclic.

Therefore, every proper subgroup of G is cyclic.

**EDIT:** made more clear.

**7.20**: Suppose H and K are subgroups of a group G. If |H| = 12 and |K| = 35, find |H∩K|. Generalize.

First we must show that |H∩K| is a subgroup of G. It is nonempty because the identity is in H and K and thus must be in |H∩K|. Let $x \in H∩K\}$, then $x \in H\}$ and $x \in K\}$, Since $H \in G\}$ and $K \in G\}$, we have that $x \in G\}$. Next let $a,b \in H∩K\}$, then $a,b \in H\}$ and $a,b \in K\}$ Since they are groups, a^{-1}, and a*b are in H and K, thus $a*b \in H∩K\}$ and $a^^-1^^ \in H∩K\}$. Therefore, |H∩K| is a subgroup of G. So by lagranges theorem, we know that |H∩K| must divide |H| and |K|. The GCD(12,35) = 1, so |H∩K| = 1, and H∩K = {e}.

**7.21**: Suppose that $H$ is a subgroup of $S_4$ and that $H$ contains $(12)$ and $234$. Prove that $H=S_4$.

The first thing we must notice is that $H$ is a group and thus must have group properties. This will mean that every element of $H$ will have an inverse, and that that composition of any two elements of $H$ will also be in $H$. We will use these two facts to try to construct the $stab_H(1)$.

First, $(234)(12)=(1342)$ is in $H$. $(1342^{-1})=(3124)$, so $(3124)$ is also in $H$. But $(3124)$ must have an inverse itself, which is $(3421)$ (note that this is just another way of writing $1342$, so these two elements are inverses of each other and will keep alternating if we keep finding inverses). So $H$ must have $(1342)$ and $(3124)$.

By the same logic, if we start with $(234)$ and keep taking inverses, will we get that $(234)$ and $(324)$ are in $H$.

Now we use some closure properties. Note that the composition of any of the elements we have found thus far will give another element in $H$. Because $(234)=(423)=(42)(23)$, we compose $(1342)(234)(234)=(13)(34)(42)(42)(23)(23)(34)=(13)$, so $(13)$ in $H$.

Now $(13)(12)(3142)=(13)(31)(12)(12)(24)=(24)$, so $24$ is in $H$.

Then $(13)(1342)(24)=(13)(13)(34)(42)(24)=(34)$, so $(34)$ is in $H$.

Finally, $(24)(423)=(24)(42)(23)=(23)$, so $(23)$ is in $H$.

So thus far, $stab_H(1)=(234),(324),(23),(34),(42),(e)$ (because $H$ is a subgroup and therefore must have an identity.) We can also note from all the above permutations that we've shown are in $H$ that $orbit_H(1)={1,2,3,4}$. So by the orbit stabilizer theorem, $\lvert H \rvert=24$. We note that all of the $24$ elements in $H$ must be in $S_4$, otherwise $H$ would not be a subgroup of $S_4$. Therefore, $H=S_4$

**7.27**: Let $|G|=33$. What are the possible orders for the elements of $G$? Show that $G$ must have an element of order 3.

From Corollary 2 of Lagrange's Theorem, all the elements of $G$ must have orders that divide $G$. This means the order of any given element is either of order 1, 3, 11, or 33. Now lets show there exists an element of order 3.

There is only one element of order 1, the identity, this leaves 32 other elements of $G$ that don't have order 1. If one of the elements, labeled $a$, has order 33, then $a^{11}$ would have order 3 since $(a^{11})^{3}=a^{33}=e$ and we are done. The other case would be if there was no element of order 33, which leaves us with 11 and 3 as the remaining possible orders for the 32 elements. Since we'd be done if one of these elements had order 3, we must only show that not all the the remaining elements can have order 11.

Let element $b$ have order 11, then $<b>$ has order 11 and is a subgroup of $G$ by Theorem 3.4. Since 11 is prime, this subgroup is composed of the identity and 10 elements of order 11 by Lagrange's Theorem, Corollary 2. Since $<b>$ is a subgroup of $G$ this means there are 10 elements of $G$ that are expressible as powers of $b$ with order 11. In other words, when there is an element of order 11 is G, then there has to be another 9 elements that go along with it from the generated subgroup. This means elements of order 11 come in groups of 10. Since we said we had 32 remaining elements, and 32 isn't divisible by 10, they can't all be of order 11.

Therefore, $G$ Has an element of order 3.

**7.29**: Can a group of order $55$ have exactly $20$ elements of order $11$? Give a reason for your answer.

Yes. By the second corollary to Lagrange's Theorem, the order of each element of a finite group divides the order of the group. Noting that the order of $G$ is 55 so $G$ must be finite, the possible element orders of $G$ are 1, 55, 5, 11. Therefore, $G$ might have at least one element of order 11. By the corollary to Theorem 4.4, the number of elements of order $11$ must be divisible by the number of integers less than $11$ and relatively prime to $11$. Because 11 is prime, there are 10 integers less than 11 and relatively prime to 11 (the integers 1-10). Therefore, there could be $20$ elements of order $11$ in $G$, because 20 is divisible by 10.

**7.35:**Let $G$={(1), (12)(34), (1234)(56), (13)(24), (1432)(56), (56)(13), (14)(23), (24)(56)}

a. Find the stabilizer of 1 and the orbit of 1.

b. Find the stabilizer of 3 and the orbit of 3.

c. Find the stabilizer of 5 and the orbit of 5.

The stabilizer of an element in G includes all permutations so that the element is sent to itself.

The orbit of an element in G includes all possible elements in G the given element can be sent to.

a. $stab$_{$G$}(1) = {(1), (24)(56)}

$orb$_{$G$}(1) = {1, 2, 3, 4}

b. $stab$_{$G$}(3) = {(1), (24)(56)}

$orb$_{$G$}(3) = {3, 4, 1, 2}

b. $stab$_{$G$}(5) = {(1), (12)(34), (13)(24), (14)(23)}

$orb$_{$G$}(5) = {5, 6}

**7.37**: Prove that the eight-element set in the proof of Theorem 7.4 is a group.

The set mentioned is $A={e,a,a^2,a^3,b^2,b^2a,b^2a^2,b^2a^3}$, where $a=(1234)$ and $b=(1423)$. To show that A is a group, we must show that every element has an inverse, there is an identity element, the set is closed under function composition, and the set is associative.

There is an identity element by definition; $e$ is the identity permutation and it is clearly in the set.

Function composition is associative (see, for example, Gallian p. 20, Theorem 0.7-1.) and we want to show the set is a group under function composition, so associativity is satisfied.

To help us think about inverses and closure, we first calculate:

$a=(1234)$

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

$a^3=(1432)$

$a^4=e$

So the order of $a$ is 4.

For $b$:

$b^2=(12)(34)$

$b^4=e$

In addition:

$b^2a=(13)$

$b^2a^2=(14)(23)$

$b^2a^3=(24)$

From this we can see that the inverse of $a$ is $a^3$ (and the other way around); the inverse of $a^2$ is itself; and the inverse of $e$ is itself as well. The inverse of $b^2a$ and $b^2a^3$ are also themselves because the inverse of any two-cycle is itself. Also, the inverse of $b^2a^2$ is itself.

Finally, we must show that the set is closed. We note that any composition of elements will just be a combination of powers of $a$ and even powers of $b$. Because both $a$ and $b$ both have order 4, every composition will be a composition of two of the 7 elements listed above. So we just check to make sure these are closed:

$b^2a=(13)=b^2a$

$b^2a^2=(14)(23)=b^2a^2$

$b^2a^3=(24)=b^2a^3$

$b^2a^4=(12)(34)=b^2$

$b^4a==a$

$b^4a^2=a^2$

$b^4a^3=a^3$

$b^4a^4=e$

Therefore, because every composition of two or more elements can be written as one of the above combinations and because every one of the above combinations gives back an elements in the set, the set is closed.

Therefore, $A$ is a group.

**7.44:** Let G be the group of rotations of a plane about a point P in the plane. Thinking of G as a group of permutations of the plane, describe the oribit of a point Q in the plane. (This is the motivation for the name "orbit.")

We must think of P as being a fixed point in the plane such that P $\leq$ Q. If P were equal to Q, then the orbit of Q would only have Q in it such that Q is not moving, so Q would fix itself.

Suppose P $\leq$ Q , then P is a fixed point on the plane such that Q is rotating around it. This rotation of $\theta$ each time creates a circle has a center of P and radius PQ for any point Q in the rotation. This is because for every $\theta$ the symmetry is preserved so Q is equidistant to P no matter the angle and therefore we see that the orbit of a point Q is every point Q on the plane equidistant to P.

**7.45:** Let $G$ be the rotation group of a cube. Label the faces of the cube 1 through 6, and let $H$ be the subgroup of elements of $G$ that carry face 1 to itself. If $\sigma$ is a rotation that carries face 2 to face 1, give a physical description of the coset $H\sigma$.

Let line $l$ be the line perpendicular through face 1, and let $R$_{$n$}$l$ be the rotation around $l$ with $n$ degrees.

$H=stab$_{$G$}$(face 1)=\{e,R$_{$90$}$l,R$_{$180$}$l,R$_{$270$}$l\}$

Let line $m$ be the line parallel to face 2 and perpendicular to line $l$ in the center of the cube, as well as perpendicular to any face adjoining face 2.

Since we don't know the location of face 2, we can assume that it is equivalent to one of the five remaining faces: either one of the four faces adjoining face 1, or the face opposite face 1.

Let's assume that face 2 is one of the faces adjoining face 1.

Then $\sigma =R$_{$90$}$m$ in the direction that sends face 2 to face 1.

Let's assume that line $m$ remains parallel to face 2 no matter how $H$ rotates around

face 1.

If we rotate around line $l$ in any way, and then rotate around line $m$, face 2 should still go to face 1. Therefore, $H\sigma =\sigma$.

Let's assume that line $m$ remains fixed in 3D space, regardless of what face it is parallel to it.

Only one element of $H$ after executed can lead $\sigma$ to move face 2 to face 1, and that is $e$. Every other rotation of $H$ will lead $\sigma$ to face a different face.

Now, let's go ahead and assume that face 2 is opposite face 1 on the cube.

Line $m$ is still parallel to face 2 and perpendicular to line $l$. Except now, line $l$ is also perpendicular to face 2 and line $m$ is also parallel to face 1.

A quick spin of the cube will bring face 2 to face 1, so $\sigma =R$_{$180$}$m$.

At this point, it doesn't matter if line $m$ moves or is fixed according to the rotations of $H$, since a rotation of $H$ does not change the position of face 2.

Therefore, in the terms of which face faces us, $H\sigma =H$.

**7.48: a&b**

Calculate the orders of the following:

a. The group of rotations of a regular tetrahedron (a solid with four congruent equilateral triangles as faces)

b. The group of rotations of a regular octahedron (a solid with eight congruent equilateral triangles as faces)

According to the orbit-stabilizer theorem, |G| = |$orb$_{$G$}(i)| |$stab$_{$G$}(i)|

a. G = rotations of a regular tetrahedron

$orb$_{$G$}(1) = {T_{1} , T_{2} , T_{3} , T_{4}}

$stab$_{$G$}(1) = {Id, R_{120}, R_{240}}

|$orb$_{$G$}(1)| = 4

|$stab$_{$G$}(1)| = 3

|G| = (4)(3) = 12

b. G = rotations of a regular octahedron

$orb$_{$G$}(1) = {T_{1} , T_{2} , T_{3} , T_{4} , T_{5} , T_{6} , T_{7} , T_{8}}

$stab$_{$G$}(1) = {Id}

|$orb$_{$G$}(1)| = 8

|$stab$_{$G$}(1)| = 1

|G| = (8)(1) = 8

**7.48: c&d**

Calculate the orders of the following:

c. The group of rotations of a regular dodecahedron (a solid with twelve congruent regular pentagons as faces)

d. The group of rotations of a regular icosahedron (a solid with 20 congruent equilateral triangles as faces)

According to the orbit-stabilizer theorem, |G| = |$orb$_{$G$}(i)| |$stab$_{$G$}(i)|

c. G = rotations of a regular dodecahedron

$orb$_{$G$}(1) = {P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}, P_{8}, P_{9}, P_{10}, P_{11}, P_{12}}

$stab$_{$G$}(1) = {Id, R_{72}, R_{144}, R_{216}, R_{288}}

|$orb$_{$G$}(1)| = 12

|$stab$_{$G$}(1)| = 5

|G| = (12)(5) = 60

d. G = rotations of a regular icosahedron

$orb$_{$G$}(1) = {T_{1} , T_{2} , T_{3} , … , T_{19}, T_{20}}

$stab$_{$G$}(1) = {Id, R_{120}, R_{240}}

|$orb$_{$G$}(1)| = 20

|$stab$_{$G$}(1)| = 3

|G| = (20)(3) = 60

**12.2:** The ring {0,2,4,6,8} under addition and multiplication modulo 10 has a unity. Find it.

In order to find the unity in this ring, we must look for the multiplicative identity of {0,2,4,6,8}. To find this identity, we can look at the Cayley Table of {0,2,4,6,8} under multiplication mod 10:

0 | 2 | 4 | 6 | 8 | |
---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 |

2 | 0 | 4 | 8 | 2 | 6 |

4 | 0 | 8 | 6 | 4 | 2 |

6 | 0 | 2 | 4 | 6 | 8 |

8 | 0 | 6 | 2 | 8 | 4 |

From this Cayley Table, we can see that the identity is 6, since both the row and column for 6 both give us the original elements {0,2,4,6,8} in order. Therefore, for the ring {0,2,4,6,8} under addition and multiplication modulo 10, we have a unity of 6.

**29.1:**

Determine the number of ways in which the four corners of a square can be colored with two colors. (It is permissable to use a single color on all four corners.)

There are 6 ways.

Let the colors be black and white. (Assume a corner that is not black is white.) We can have all corners white; all corners black; one corner black; three corners black; two corners on adjacent sides of the square black; two corners diagonally opposite each other black. These are the six possibilities, and there are not any more than these 6, because the orbit of any other possibility would overlap the orbit of one of those six.

**29.2:**

Determine the number of different necklaces that can be made using 13 white beads and three black beads.

In order to find the number of different necklaces, we can think of finding the number of orbits and therefore, can use Burnside's Theorem $\frac{1}{|G|} \sum |fix(\theta)|$. Since we have 16 beads in total, we can think of this as a 16 sided polygon or $D_{16}$. We know that |$D_{16}$| =32 by the fact that it is a dihedral group, so this corresponds to our |G|. We also know that there are ${16 \choose 13}$ = 560 arrangements of the beads. Therefore, there are 560 ways that e is fixed so |fix(e)| =560. We now must look at the different types of symmetry of this 11 sided polygon. These symmetries are: rotations, reflections through beads, and reflections between beads.

When looking at rotations (other than the identity) we know that our rotations are all multiples of 360/11 and we have 11 of them. But, all of the rotations except for the identity will not be able to fix itself since when we rotate the necklace, no matter the degrees, we will not be able to result in a necklace with 13 white beads and three black beads in the positions we need them to be such that a white moves to a white in every rotation to make a necklace that has the right amount of colored beads.

Now we must look at the reflections through the beads, which we have 8 of. These lines of reflections have 7 beads on either side of the line of reflection. Then, to figure out the order of the fix set of each of these, we know we have 7 arrangements of white beads on either side of the reflection and we have 2 spots on the line of reflection that get the white bead, and therefore we have 7x2=14 as the order of the fix set one of these reflections. And since we have 8 of these reflection lines, we have 8x14=112 as the order of all the fix sets of these types of reflections.

Lastly, we must look at the reflections between beads. These reflections split the beads in half with 8 beads being on one side of the line and 8 beads being on the other side of the line. We know that the order of the fix set of each of these reflections is 0 because we can place two of the white beads symmetrical around the line of reflection, but the 3rd white bead must reflect to a black bead.

Then using Burnside's Theorem, we get $\frac{1}{32} (560+112)$ = 672/32 = 21. Therefore, there are 21 different necklaces using 13 white beads and 3 black beads.

**29.3:**

Determine the number of ways in which the vertices of an equilateral triangle can be colored with five colors so that at least two colors are used.

The triangle is equilateral, so each corner of the triangle looks exactly the same as any other triangle, making making which corner which in some sense irrelevant since you can rotate or flip the triangle around to make any corner be able to switch with any other corner. In class we showed the triangle is isomorphic to $S_{3}$, so we really can make any permutation of the 3 corners we want.

Since we need at least 2 colors, and at most 3 colors, we can break this up into two cases, where we are using 3 distinct colors and when we are using 2 distinct colors. First, if we are using 3 colors of the 5, we have 5 ways to color the first corner, 4 ways for the second, and 3 ways for the last. This would suggest 60 ways to color the triangle in this case alone. But, we don't care what order those colors come in because we can just permute the corners to some different order. In other words, we can chose blue, white, then red, or we can chose red, white, then blue, but there really is no difference. The number of ways we can arrange 3 colors is 6, so we must divide 60 by 6 to get 10 distinct coloring in this case.

For the second case using two colors, we have 5 says of choosing the color we plan to use twice, then 4 ways of choosing the color we will use once. This gives 20 combinations. This choice really does matter because choosing to use red twice and then blue once is different than choosing to use blue twice and then red once. Once we have our colors chosen it doesn't matter how we paint the triangle, we can just flip it around to any of the possible corner orders.

When you add the two cases you are left with 30 combinations.