Test 2 Review
Definitions:
Permutation, odd permutation, even permutation, isomorphism, automorphism, coset, orbit, stabilizer, normal.
Groups:
$\Bbb Q,\Bbb Q^*,\Bbb R,\Bbb R^*,\Bbb C,\Bbb C^∗,\Bbb Z,\Bbb Z_n,U(n),D_n,S_n,A_n$
Theorems:
Fundamental Theorem of Cyclic Groups, $\operatorname{Aut}(\Bbb Z_n) ≈ U(n)$, Lagrange’s Theorem, Orbit-Stabilizer Theorem.
Know how to:
prove a group is cyclic, prove something is a subgroup, prove something is an isomorphism, prove two groups are isomorphic, prove two groups are not isomorphic, compute left and right cosets, compute products and inverses of permutations.
Know the proof of Lagrange’s Theorem.

Test 2 Solutions November 2, 2011
1. State and prove Lagrange's theorem; give full details. Use this result to prove that every group of prime order is cyclic.
Lagrange's theorem states that for any subgroup $H$ of a finite group $G$, the order of $H$ divides the order of $G$. In fact, the quotient $|G|/|H|$ is equal to the number of left (or right) cosets of $H$ in $G$.
Suppose $g_{1} H, g_{2} H, \ldots, g_{n} H$ are the distinct left cosets of $H$ in $G$. We know that all of these cosets have the same size, which is equal to $|H|$ (since $H$ is, in fact, one of these cosets). Moreover, they partition $G$; i.e., they are pairwise disjoint and every element of $G$ is in one of them. Thus $|G|=n|H|$.
2. (i) Define the terms permutation, odd permutation, and even permutation. Define the groups $S_n$ and $A_n$ and prove that $A_n\leq S_n$.
A permutation is a one-to-one map from a set onto itself. A permutation is even if it has an expression as a product of an even number of 2-cycles. It is odd if it has an expression as a product of an odd number of 2-cycles.
(ii) Let $\alpha=(12)(54)$ and $\beta=(16532)$. Compute $\beta^{-1}, \alpha \beta$, and $\beta \alpha$. (Your answers should be in disjoint cycle notation.)
The inverse of $\beta$ is $\beta^{-1}=(23561)$ or any cyclic permutation of this. The product $\alpha \beta$ is equal to $\alpha \beta=(12)(54)(16532)=(16453)$. The product $\beta \alpha$ is equal to $\beta \alpha=(16532)(12)(54)=(26543)$.
3. (i) Complete the following definition: An isomorphism $\theta: G \rightarrow \bar{G}$ is ⋯
⋯ a one-to-one map from $G$ onto $\bar{G}$ that preserves the operations. In other words, $\theta(g h)=\theta(g) \theta(h)$ for all $g, h \in G$, where the operation on the left is the product in $G$ and the operation on the right is the product in $\bar{G}$.
(ii) Recall that an automorphism of $G$ is an isomorphism $G \rightarrow G$. Prove that if $\alpha$ and $\beta$ are automorphisms of $G$, then so is $\alpha \beta$.
One-to-one: Suppose $\alpha \beta(x)=\alpha \beta(y)$. Then $\alpha(\beta(x))=\alpha(\beta(y))$. Thus the fact that $\alpha$ is one-to-one implies that $\beta(x)=\beta(y)$. Then the fact that $\beta$ is one-to-one implies that $x=y$. Thus $\alpha \beta$ is one-to-one.
Onto: Pick $g \in G$. Since $\alpha$ is onto, there is some $g^{\prime} \in G$ such that $\alpha\left(g^{\prime}\right)=g .$ Since $\beta$ is onto, there is some $g^{\prime \prime} \in G$ such that $\beta\left(g^{\prime \prime}\right)=$ $g^{\prime}$. Then $\alpha \beta\left(g^{\prime \prime}\right)=\alpha\left(g^{\prime}\right)=g$. Thus $\alpha \beta$ is onto.
Operation-preserving: Since $\beta$ is operation-preserving, we know that $\beta(g h)=\beta(g) \beta(h)$. Since $\alpha$ is operation-preserving, we know that $\alpha(\beta(g) \beta(h))=$ $\alpha(\beta(g)) \alpha(\beta(h))$. Thus $\alpha \beta(g h)=\alpha \beta(g) \alpha \beta(h)$, so that $\alpha \beta$ is operation-preserving.
(iii) Let $g \in G$ and define $T_{g}: G \rightarrow G$ by $T_{g}(h)=g^{-1} h g$ for all $h \in G$. Prove that $T_{g}$ is an automorphism of $G$.
Suppose $T_{g}(h)=T_{g}\left(h^{\prime}\right)$. Then $g^{-1} h g=g^{-1} h^{\prime} g$. Multiplying both sides by $g$ on the left and $g^{-1}$ on the right, we find that $h=h^{\prime}$, so that $T_{g}$ is one-to-one.
Fix $h \in G$, and consider $h^{\prime}=g h g^{-1}$. Then $T_{g}\left(h^{\prime}\right)=g^{-1} h^{\prime} g=g^{-1} g h g^{-1} g=$ $h$, so that $T_{g}$ is onto.
(Alternatively, one can simply observe that $T_{g^{-1}}=T_{g}^{-1}$ to see that $T_{g}$ is one-to-one and onto.)
Finally, note that $T_{g}(a b)=g^{-1} a b g=g^{-1} a g g^{-1} b g=T_{g}(a) T_{g}(b)$, so that $T_{g}$ is operation-preserving.
4. Prove or disprove:
(i) $\mathbf{R}^{*} \approx \mathbf{C}^{*}$.
The equation $x^{4}=1$ has exactly four roots in $\mathbf{C}^{*}$. If $\varphi$ is an isomorphism from $\mathbf{C}^{*}$ to $\mathbf{R}^{*}$, then $\varphi(1)=1$. Thus $\varphi$ takes solutions to $x^{4}=1$ in $\mathbf{C}^{*}$ to solutions to $x^{4}=1$ in $\mathbf{R}^{*}$. But $x^{4}=1$ has only two solutions in $\mathbf{R}^{*}$. Thus no such isomorphism can exist.
(ii) $U(10) \approx U(5)$
We have that $U(10)=\{1,3,7,9\}$ and $U(5)=\{1,2,3,4\}$. Define $\phi: U(5) \rightarrow$ $U(10)$ by $\phi(1)=1, \phi(2)=3, \phi(3)=7$, and $\phi(4)=9$. We claim that $\phi$ is an isomorphism. Clearly $\phi$ is one-to-one and onto. To see that it preserves the operations, we can either check cases individually, or observe that the Cayley tables for the groups are identical up to substitution corresponding to $\phi$.
(iii) $S_{n} \approx A_{n}$
This is actually true if $n=1$, as both groups are trivial. Thus we may assume that $n \geq 2$. Then since $A_{n}$ is a subgroup of $S_{n}$ that does not contain (12), for instance, and $S_{n}$ is a finite group, it follows that $\left|A_{n}\right|<\left|S_{n}\right|$, so that $S_{n}$ and $A_{n}$ are not isomorphic.
5. Let $G$ be a group of permutations on a set $X$. Let $x \in X$.
(i) Define $\operatorname{stab}_{G}(x)$, the stabilizer of $x$ in $G$.
We define $\operatorname{stab}_{G}(x)=\{g \in G \mid g x=x\}$.
(ii) Define $\operatorname{orb}_{G}(x)$, the orbit of $x$ under $G$.
We define $\operatorname{orb}_{G}(x)=\{y \in X \mid g x=y\text{ for some }g \in G\}$.
(iii) Suppose we define a relation $\sim$ on $X$ by saying that $a \sim b$ if and only if $\operatorname{orb}_{G}(a)=$ $\operatorname{orb}_{G}(b)$. Prove that $\sim$ is an equivalence relation.
Observe that $\operatorname{orb}_{G}(a)=\operatorname{orb}_{G}(a)$, so that $a \sim a$. Also$$\operatorname{orb}_{G}(a)=\operatorname{orb}_{G}(b) \quad \Rightarrow \quad \operatorname{orb}_{G}(b)=\operatorname{orb}_{G}(a),$$so that $a \sim b \Rightarrow b \sim a$. Finally we have$$\operatorname{orb}_{G}(a)=\operatorname{orb}_{G}(b) \quad \& \quad \operatorname{orb}_{G}(b)=\operatorname{orb}_{G}(c) \quad \Rightarrow \quad \operatorname{orb}_{G}(a)=\operatorname{orb}_{G}(c),$$so that $a \sim b$ and $b \sim c$ implies $a \sim c$. Thus $\sim$ is an equivalence relation.
6. Let $p$ be a prime and let $g \in U(p)$. Show that $g^{p-1}=1$. Hence prove that $a^{p} \equiv a\bmod p$ for each integer $a$.
Because $p$ is prime, we know that $U(p)=\{1,2, \ldots, p-1\}$, so that $|U(p)|=$ $p-1$. For any finite group $G$, we know that $g^{|G|}=e$. Thus for each $g \in U(p)$ we have $g^{p-1}=e$ in $U(p)$. Now suppose $a$ is any integer, and let $A$ be the element in $U(p)$ congruent to $a \bmod p$. Then $a^{p-1}\bmod p=A^{p-1} \bmod p=1 \bmod p$, since $a^{p-1}=1$ in $U(p)$. Multiplying both sides by $a$ gives $a^{p} \equiv a \bmod p$.
7. Let $G=\{e,(12)(34),(1234)(56),(13)(24),(1432)(56),(56)(13),(14)(23),(24)(56)\}$. Thus $G$ is a group of symmetries of the set $\{1,2,3,4,5,6\}$.
(i) Find the stabilizer of 3 and the orbit of 3 under the action of $G$.
The stabilizer of 3 is $\{e,(24)(56)\}$. The orbit is $\{1,2,3,4\}$.
(ii) Use the orbit-stabilizer theorem to compute the size of the rotational symmetry group of the regular octahedron.
Let $x$ be the center of a face. Then the orbit of $x$ has size eight, corresponding to the eight faces of the octagon. The stabilizer of $x$ has order three, corresponding to rotation of the face containing $x$. Thus the order of the rotational symmetry group is $8 \cdot 3=24$.
Alternatively, let $x$ be a vertex. Then the orbit of $x$ has size six, corresponding to the six vertices. The stabilizer has order four, corresponding to cycling through the four faces adjacent to $x$ (alternatively, think of cycling through the four edges adjacent to $x$). Thus the order of the rotational symmetry group is $6 \cdot 4=24$.
(iii) Find all right cosets of $U=\left\{R_{0}, V\right\}$ in $D_{4}=\left\{R_{0}, R_{90}, R_{180}, R_{270}, H, V, D, D^{\prime}\right\}$. Show that $U$ is not normal in $D_{4}$ by finding an element $g$ of $D_{4}$ such that $g U \neq U g$.
There are four cosets:\begin{gathered}U R_{0}=U=U VU R_{90}=\left\{R_{0} R_{90}, V R_{90}\right\}=\left\{R_{90}, D^{\prime}\right\}=\left\{D^{\prime}, R_{90}\right\}=\left\{R_{0} D^{\prime}, V D^{\prime}\right\}=U D^{\prime} \\U R_{180}=\left\{R_{0} R_{180}, V R_{180}\right\}=\left\{R_{180}, H\right\}=\left\{H, R_{180}\right\}=\left\{R_{0} H, V H\right\}=U H \\U R_{270}=\left\{R_{0} R_{270}, V R_{270}\right\}=\left\{R_{270}, D\right\}=\left\{D, R_{270}\right\}=\left\{R_{0} D, V D\right\}=U D\end{gathered}Observe that $R_{90} U=\left\{R_{90}, D\right\} \neq U R_{90}$. Thus $U$ is not normal.