## Tag Archives: automorphism

### The order of the Frobenius map on a finite field

Let $\varphi$ denote the Frobenius map $x \mapsto x^p$ on $\mathbb{F}_{p^n}$. Prove that $\varphi$ is an automorphism and compute its order in $\mathsf{Aut}(\mathbb{F}_{p^n})$.

Recall that $\varphi$ is a homomorphism. Moreover, if $\alpha \in \mathsf{ker}\ \varphi$, then $\alpha^p = 0$. Since fields contain no nontrivial zero divisors, we have $\alpha = 0$ (using induction if you want). So the kernel of $\varphi$ is trivial, and thus $\varphi$ is injective. Since $\mathbb{F}_{p^n}$ is finite, $\varphi$ is surjective, and so is a field isomorphism.

Next, we claim that $\varphi^t(\alpha) = \alpha^{p^t}$ for all $\alpha$ and all $t \geq 1$, and will show this by induction. The base case certainly holds, and if $\varphi^t(\alpha) = \alpha^{p^t}$, then $\varphi^{t+1}(\alpha) = \varphi(\varphi^t(\alpha)) = \varphi(\alpha^{p^t})$ $= (\alpha^{p^t})^p$ $= \alpha^{p^{t+1}}$ as desired.

Now $\varphi^n(\alpha) = \alpha^{p^n} = \alpha$, since the elements of $\mathbb{F}_{p^n}$ are precisely the roots of $x^{p^n}-x$. So we have $\varphi^n = 1$.

If $\varphi^t = 1$, then we have $\alpha^{p^t} - \alpha = 0$ for all $\alpha$, so that each $\alpha$ is a root of $x^{p^t}-x$. So $x^{p^n-1}-1$ divides $x^{p^t-1}-1$, and so $p^n-1$ divides $p^t-1$ (by this previous exercise) and then $n$ divides $t$ (by this previous exercise). In particular, $n \leq t$.

So $n$ is the order of $\varphi$ in $\mathsf{Aut}(\mathbb{F}_{p^n})$.

### Exhibit a quadratic field as a field of matrices

Let $K = \mathbb{Q}(\sqrt{D})$, where $D$ is a squarefree integer. Let $\alpha = a+b\sqrt{D}$ be in $K$, and consider the basis $B = \{1,\sqrt{D}\}$ of $K$ over $\mathbb{Q}$. Compute the matrix of the $\mathbb{Q}$-linear transformation ‘multiplication by $\alpha$‘ (described previously) with respect to $B$. Give an explicit embedding of $\mathbb{Q}(\sqrt{D})$ in the ring $\mathsf{Mat}_2(\mathbb{Q})$.

We have $\varphi_\alpha(1) = a+b\sqrt{D}$ and $\varphi(\alpha)(\sqrt{D}) = bD + a\sqrt{D}$. Making these the columns of a matrix $M_\alpha$, we have $M_\alpha = \begin{bmatrix} a & bD \\ b & a \end{bmatrix}$, and this is the matrix of $\varphi_\alpha$ with respect to $B$. As we showed in the exercise linked above, $\alpha \mapsto M_\alpha$ is an embedding of $K$ in $\mathsf{Mat}_2(\mathbb{Q})$.

Compare to this previous exercise about $\mathbb{Z}[\sqrt{D}]$.

### Prove that a given map is a field automorphism

Prove directly that the map $\varphi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$ given by $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ is a field isomorphism.

Note that elements of $\mathbb{Q}(\sqrt{2})$ can be written uniquely in the form $a+b\sqrt{2}$ where $a,b \in \mathbb{Q}$, so that this rule indeed defines a function.

Now $\varphi((a+b\sqrt{2}) + (c+d\sqrt{2})) = \varphi((a+c) + (b+d)\sqrt{2})$ $= (a+c) - (b+d)\sqrt{2}$ $= (a-b\sqrt{2}) + (c-d\sqrt{2})$ $= \varphi(a+b\sqrt{2}) + \varphi(c+d\sqrt{2})$, so that $\varphi$ preserves addition.

Similarly, we have $\varphi((a+b\sqrt{2})(c+d\sqrt{2})) = \varphi((ac+2bd) + (ad+bc)\sqrt{2})$ $= (ac+2bd) - (ad+bc)\sqrt{2}$ $= (a-b\sqrt{2})(c-d\sqrt{2})$ $= \varphi(a+b\sqrt{2})\varphi(c+d\sqrt{2})$, so that $\varphi$ preserves multiplication.

Thus $\varphi$ is a ring homomorphism, and hence a field homomorphism.

Note that if $\varphi(a+b\sqrt{2}) = 0$, then we have $a-b\sqrt{2} = 0$, and so if $b = 0$ we have $a/b = \sqrt{2}$, where $a$ and $b$ are rational numbers- a contradiction since $\sqrt{2}$ is not rational. So $b = 0$, and thus $a = 0$, and we have $\mathsf{ker}\ \varphi = 0$. So $\varphi$ is injective.

Finally, $\varphi$ is surjective since we have $\varphi(a+(-b)\sqrt{2}) = a+b\sqrt{2}$ for all $a$ and $b$.

So $\varphi$ is a field isomorphism.

### An automorphism of Sym(6) that is not inner

This exercise exhibits an automorphism of $S_6$ that is not inner (hence it shows that $[\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] = 2$. Let $u_1 = (1\ 2)(3\ 4)(5\ 6)$, $u_2 = (1\ 4)(2\ 5)(3\ 6)$, $u_3 = (1\ 3)(2\ 4)(5\ 6)$, $u_4 = (1\ 2)(3\ 6)(4\ 5)$, and $u_5 = (1\ 4)(2\ 3)(5\ 6)$. Show that $u_1, \ldots, u_5$ satisfy the following relations:

• $u_i^2 = 1$ for all $i$,
• $(u_iu_j)^2 = 1$ for all $i,j$ with $|i-j| \geq 2$, and
• $(u_iu_{i+1})^3 = 1$ for all $i \in \{1,2,3,4\}$.

Deduce that $S_6 = \langle u_1, \ldots, u_5 \rangle$ and that the map $(1\ 2) \mapsto u_1$, $(2\ 3) \mapsto u_2$, $(3\ 4) \mapsto u_3$, $(4\ 5) \mapsto u_4$, $(5\ 6) \mapsto u_5$ extends to an automorphism of $S_6$ (which is clearly not inner since it does not preserve cycle shape.

Showing that $u_i$ satisfy the given relations is a straightforward computation. Moreover, they generate $S_6$ since $u_3u_5u_1 = (6\ 5)$ and $u_1u_2u_4 = (6\ 5\ 2\ 3\ 4\ 1)$.

### Count the number of automorphisms of a group from a presentation

Use presentations to find the orders of the automorphism groups of $Z_2 \times Z_4$ and $Z_4 \times Z_4$.

We have $Z_2 \times Z_4 = \langle a,b \ |\ a^2 = b^4 = 1, ab = ba \rangle$.

Now any element $x$ of order 4 and any element $y$ of order 2 with $x \neq y^2$ generate $Z_2 \times Z_4$. There are 4 elements of order 4: $b$, $b^3$, $ab$, and $ab^3$. There are 2 elements of order 2: $a$, $b^2$, and $ab^2$. Once $x$ is chosen, there are two choices for $y$, and each choice determines an automorphism. These are distinct by construction. Thus $|\mathsf{Aut}(Z_2 \times Z_4)| = 8$.

Now $Z_4 \times Z_4 = \langle a,b \ |\ a^4 = b^4 = 1, ab = ba \rangle$.

Any two elements of order 4, say $x$ and $y$, generate $Z_4 \times Z_4$ provided $\langle x \rangle \cap \langle y \rangle = 1$. This group has 12 elements of order 4, and these intersect nontrivially pairwise. Thus once $x$ is chosen, there are 8 choices for $y$. Thus $|\mathsf{Aut}(Z_4 \times Z_4)| = 96$.

### Given an automorphism of order 2 of an elementary abelian group, the set of fixed points has bounded rank

Let $V$ be an elementary abelian 2-group of rank $r \geq 1$, and let $\varphi$ be an automorphism of $V$ such that $\varphi^2 = 1$. Prove that the map $\psi(v) = v + \varphi(v)$ is a homomorphism of $V$ to itself and that every element in the image of $\psi$ is fixed by $\varphi$. Deduce that the set of elements fixed by $\varphi$ is an elementary abelian subgroup of rank $t$, where $2t \geq r$.

First we show that $\psi$ is a homomorphism. If $v,w \in V$, then $\psi(v+w) = v+w + \varphi(v+w)$ $= v + \varphi(v) + w + \varphi(w)$ $= \psi(v) + \psi(w)$.

Now let $F \subseteq V$ be the set of elements fixed by $\varphi$; it is clear that $F$ is in fact a subgroup of $V$.

Now if $v \in \mathsf{ker}\ \psi$, then $0 = \psi(v) = v+\varphi(v)$, so that $v = \varphi(v)$. Thus $v \in F$. Likewise, if $v \in \mathsf{im}\ \psi$, then $v = w + \varphi(w)$ for some $w$, and we have $\varphi(v) = \varphi(w) + \varphi^2(w)$ $= w + \varphi(w) = v$. Thus $v \in F$.

By the First Isomorphism Theorem, we have $[V, \mathsf{ker}\ \psi] = |\mathsf{im}\ \psi|$, and since everything is finite, $|V| = |\mathsf{im}\ \psi| \cdot |\mathsf{ker}\ \psi| \leq |F|^2$.

Now $|V| = 2^r$ and $|F| = 2^t$, so that $r \leq 2t$, as desired.

### Every automorphism of an elementary abelian p-group having p-power order has nontrivial fixed points

Let $p$ be a prime, let $V$ be a nontrivial elementary abelian $p$-group, and let $\varphi$ be an automorphism of $V$ of $p$-power order. Prove that $\varphi$ has nontrivial fixed points.

Note that $\langle \varphi \rangle \leq \mathsf{Aut}(V)$; use the inclusion map to construct $P = V \rtimes \langle \varphi \rangle$. Now $P$ is a $p$-group, and $V \leq P$ is normal, so that by Theorem 1, $V \cap Z(P)$ is nontrivial.

Now let $x \in V \cap Z(P)$ with $x \neq 1$. Then $(x, \varphi) = (x,1)(1,\varphi)$ $= (1,\varphi)(x,1)$ $= (\varphi(x),\varphi)$. Comparing entries we see that $x = \varphi(x)$.

### For all n except 6, every automorphism of Sym(n) is inner

This exercise shows that for $n \neq 6$ every automorphism of $S_n$ is inner. Fix an integer $n \geq 2$ with $n \neq 6$.

1. Prove that every automorphism of a group $G$ permutes the conjugacy classes of $G$. That is, if $K \subseteq G$ is a conjugacy class and $\varphi$ an automorphism of $G$, then $\varphi[K]$ is a conjugacy class of $G$.
2. Let $K$ be the conjugacy class of transpositions in $S_n$ and let $L$ be the conjugacy class containing any element of order 2 in $S_n$ that is not a transposition. Prove that $|K| \neq |L|$. Deduce that any automorphism of $S_n$ sends transpositions to transpositions. [Hint: Use §4.3 #33.]
3. Prove that for each $\sigma \in \mathsf{Aut}(S_n)$, $(1\ 2) \mapsto (a\ b_2)$, $(1\ 3) \mapsto (a\ b_3)$, …, $(1\ n) \mapsto (a\ b_n)$ for some distinct integers $a, b_2, \ldots, b_n \in \{1,2,\ldots,n\}$.
4. Show that $(1\ 2)$, $(1\ 3)$, …, $(1\ n)$ generate $S_n$ and deduce that any automorphism of $S_n$ is uniquely determined by its action on these elements. Use (c) to show that $S_n$ has at most $n!$ automorphisms. Conclude that $\mathsf{Aut}(S_n) = \mathsf{Inn}(S_n)$ for $n \neq 6$.

1. [Short way]

Let $\sigma$ be an automorphism of $G$, and let $G \cdot a$ be a conjugacy class. Note that $\sigma(g \cdot a) = \sigma(g^{-1}ag)$ $= \sigma(g) \cdot \sigma(a)$, so that $\sigma[G \cdot a] = \sigma[G] \cdot \sigma(a)$ $= G \cdot \sigma(a)$. In particular, $\sigma[G \cdot a]$ is a conjugacy class of $G$.

[Long way]

Let $\sigma \in \mathsf{Aut}(G)$ and let $K \subseteq G$ be a conjugacy class. We claim that $\sigma[K]$ is a conjugacy class. Let $a,b \in \sigma[K]$. Then $a = \sigma(x)$ and $b = \sigma(y)$ for some $x,y \in K$. Since $K$ is a conjugacy class, there exists an element $z \in G$ such that $zxz^{-1} = y$; then $\sigma(z)\sigma(x)\sigma(z)^{-1} = \sigma(y)$, so we have $\sigma(z) a \sigma(z)^{-1} = b$. Thus $a$ and $b$ are conjugate.

Let $a \in \sigma[K]$ and $b \notin \sigma[K]$ and suppose that $a$ and $b$ are conjugate; then for some $c \in G$, $cac^{-1} = b$. Now since $\sigma$ is surjective, we have $a = \sigma(x)$, $b = \sigma(y)$, and $c = \sigma(z)$ for some $x \in K$, $b \notin K$, and $c \in G$. Now $\sigma(zxz^{-1}) = \sigma(y)$, and since $\sigma$ is injective, we have $zxz^{-1} = y$. Thus $x$ and $y$ are conjugate, a contradiction since $K$ is a conjugacy class.

Thus $\sigma[K]$ is a conjugacy class for all automorphisms $\varphi$ and conjugacy classes $K$. Moreover, we have $\mathsf{id}[K] = K$ and $(\varphi \circ \psi)[K] = \varphi[\psi[K]]$, so that $\mathsf{Aut}(G)$ acts on the set of conjugacy classes of $G$ by function application.

2. We now prove some lemmas.

Lemma 1: If $k \geq 4$, then $\frac{2(2k-2)!}{k! \cdot 2^k} > 1$. Proof: We proceed by induction on $k$. For the base case ($k = 4$) note that $\frac{2(8-2)!}{4! \cdot 2^4} = \frac{2 \cdot 6!}{4! \cdot 2^4} = \frac{6 \cdot 5}{8} = \frac{15}{4} > 1$. For the inductive step, suppose that the conclusion holds for some $k \geq 4$. Now note that the polynomial $p(k) = 4k^2 - 4k - 2$ has roots $\frac{1 \pm \sqrt{3}}{2}$ and opens upward; thus for $k \geq 4$, $p(k) > 0$. Now $4k^2 - 2k > 2k + 2$, so that $\frac{2k(2k-1)}{2(k+1)} > 1$. Thus $1 < \frac{2(2k-2)!}{k! \cdot 2^k} \cdot \frac{2k(2k-1)}{2(k+1)} = \frac{2(2(k+1) - 2)!}{(k+1)! \cdot 2^{k+1}}$. By induction, the conclusion holds for all $k \geq 4$. $\square$

Lemma 2: Suppose $4 \leq 2k \leq n$. If $\frac{2(n-2)!}{k! 2^k (n-2k)!} > 1$, then $\frac{2((n+1)-2)!}{k! 2^k ((n+1) - 2k)!} > 1$. Proof: Note that $1 < k$, so that $2 < 2k$. Hence $-2k + 1 < -1$, so that $n - 2k + 1 < n-1$. Thus $1 < \frac{2(n-2)!}{k! 2^k (n-2k)!} \cdot \frac{n-1}{n+1-2k} = \frac{2((n+1)-2)!}{k! 2^k ((n+1)-2k)!}$. $\square$

Lemma 3: If $n \geq 8$ and $4 \leq 2k \leq n$, then $\frac{2(n-2)!}{k! 2^k (n-2k)!} > 1$. Proof: We proceed by induction on $n$. For the base case ($n = 8$), there are three possibilities for $k$. If $k = 2$, we have $\frac{2(8-2)!}{2! 2^2 (8-4)!} = \frac{15}{2} > 1$. If $k = 3$, we have $\frac{2(8-2)!}{3! 2^3 (8-6)!} = \frac{5}{2} > 1$. For $k = 4$ the inequality holds by Lemma 1. Now for the inductive step, suppose that for some $n \geq 8$ the inequality holds for all $4 \leq 2k \leq n$. If $n$ is even, then $4 \leq 2k \leq n+1$ if and only if $4 \leq 2k \leq n$. Thus by Lemma 2, $\frac{2((n+1)-2)!}{k! 2^k ((n+1)-2k)!} > 1$ for all $4 \leq 2k \leq n+1$. If $n$ is odd, then $4 \leq 2k \leq n+1$ if and only if $4 \leq 2k \leq n$ or $2k = n+1$. In the first case, use Lemma 2 as before. In the second case, the inequality holds by Lemma 1. Thus for all $4 \leq 2k \leq n+1$, we have $\frac{2((n+1)-2)!}{k! 2^k ((n+1)-2k)!} > 1$. By induction, the inequality holds for all $n \geq 8$ and all $4 \leq 2k \leq n$. $\square$

Lemma 4: If $n \in \{ 4,5,7 \}$ and $4 \leq 2k \leq n$, then $\frac{2(n-2)!}{k! 2^k (n-2k)!} \neq 1$. Proof: If $n=4$, then $k = 2$. Now $\frac{2(4-2)!}{2! 2^2 (4-4)!} = \frac{1}{2} \neq 1$. If $n=5$, then $k=2$. Now $\frac{2(5-2)!}{2! 2^2 (5-4)!} = \frac{3}{2} \neq 1$. If $n=7$, then $k=2$ or $k=3$. If $k=2$, then $\frac{2(7-2)!}{2! 2^2 (7-4)!} = 5 \neq 1$. If $k=3$, then $\frac{2(7-2)!}{3! 2^3 (7-6)!} = 5 \neq 1$. $\square$.

Now we move to the main result. Note that if $n$ is 2 or 3, then all elements of order 2 are transpositions; thus any automorphism of $S_n$ sends transpositions to transpositions. Now if $n \geq 4$ and $\sigma$ is an element of order 2, then $\sigma$ is a product of $k$ disjoint 2-cycles. If $\sigma$ is not a transposition, then $k \geq 2$; note then that $4 \leq 2k \leq n$. By §4.3 #33, the number of transpositions in $S_n$ is $\frac{n!}{2(n-2)!}$, and the number of conjugates of $\sigma$ is $\frac{n!}{k! 2^k (n-2k)!}$. These two numbers are not equal for $n \neq 6$ and all $4 \leq 2k \leq n$, since otherwise we have $\frac{2(n-2)!}{k! 2^k (n-2k)!} = 1$, which contradicts Lemmas 3 and 4. Now if $\varphi$ is an automorphism of $S_n$ and $K \subseteq S_n$ the conjugacy class of transpositions, then $\varphi[K]$ is a conjugacy class of elements of order 2. If $\varphi[K] \neq K$, we have a contradiction since $|K| \neq |\varphi[K]|$. Thus every automorphism of $S_n$ maps transpositions to transpositions if $n \neq 6$.

3. We begin with some lemmas.

Lemma 5: Let $\varphi \in \mathsf{Aut}(G)$. Then for all $x,y \in G$, $xy = yx$ if and only if $\varphi(x)\varphi(y) = \varphi(y)\varphi(x)$. Proof: $xy = yx$ if and only if $\varphi(xy) = \varphi(yx)$ since $\varphi$ is injective, if and only if $\varphi(x)\varphi(y) = \varphi(y)\varphi(x)$ since $\varphi$ is a homomorphism. $\square$

Lemma 6: Let $\varphi \in \mathsf{Aut}(S_n)$ with $n \geq 3$. Then $|\bigcap_{k=2}^n M(\varphi((1\ k)))| = 1$. Proof: For each $k$, $\varphi((1\ k))$ is a transposition by part (b). Thus $|M(\varphi((1\ k)))| \leq 2$. Suppose $\bigcap_{k=2}^n M(\varphi((1\ k))) = \emptyset$; then for some $k \neq \ell$, we have $M(\varphi((1\ k))) \neq M(\varphi(1\ \ell))$. But then $\varphi((1\ k))$ and $\varphi((1\ \ell))$ are disjoint and thus commute, a contradiction since $(1\ k)$ and $(1\ \ell)$ do not commute. Now suppose $|\bigcap_{k=2}^n M(\varphi((1\ k)))| = 2$; since each $\varphi((1\ k))$ is a transposition, we have $\varphi((1\ k)) = \varphi((1\ \ell))$ for all $k$ and $\ell$, a contradiction since $\varphi$ is injective and (since $n \geq 3$) there exist $2 \leq k, \ell \leq n$ with $k \neq \ell$. Thus $|\bigcap_{k=2}^n M(\varphi((1\ k)))| = 1$. $\square$

Now to the main result. If $n=2$, then $S_n$ has only the trivial automorphism, under which $(1\ 2) \mapsto (1\ 2)$. If $n \geq 3$, then by part (b) each of $\varphi((1\ k))$ is a transposition for $2 \leq k \leq n$, and by Lemma 6 the $\varphi((1\ k))$ move a common element $a$. Since $\varphi$ is a bijection, the second element moved by the $\varphi((1\ k))$ are pairwise distinct. Thus for some distinct $a,b_2,\ldots,b_n \in \{ 1,2,\ldots,n\}$ we have $\varphi((1\ 2)) = (a\ b_2)$, $\varphi((1\ 3)) = (a\ b_3)$, …, $\varphi((1\ n)) = (a\ b_n)$.

4. In §3.5 #3, we saw that $S_n$ is generated by $\{(a\ a+1) \ |\ 1 \leq a < n \}$. Note that $(a\ a+1) = (1\ a)(1\ a+1)(1\ a)$. Thus $S_n = \langle \{ (1\ a) \ |\ 2 \leq a \leq n \} \rangle$, and so any automorphism of $S_n$ is uniquely determined by its action on these transpositions. By part (c), there are at most $n!$ possible choices for the images of the $(1\ k)$ under an automorphism $\varphi$ (one for each assignment of the $a,b_i$ to the integers $\{1,2,\ldots,n\})$. Thus $\mathsf{Aut}(S_n) = \mathsf{Inn}(S_n)$ for $n \neq 6$.

### Characterize the automorphisms of Cyc(n)

Let $Z_n = \langle \alpha \rangle$ be a cyclic group of order $n$ and for each integer $a$, define $\sigma_a : Z_n \rightarrow Z_n$ by $\sigma_a(x) = x^a$.

1. Prove that $\sigma_a$ is an automorphism of $Z_n$ if and only if $a$ and $n$ are relatively prime.
2. Prove that $\sigma_a = \sigma_b$ if and only if $a \equiv b$ mod $n$.
3. Prove that every automorphism of $Z_n$ is equal to $\sigma_a$ for some $a$.
4. Prove that $\sigma_a \circ \sigma_b = \sigma_{ab}$. Deduce that the mapping $\overline{a} \rightarrow \sigma_a$ is an isomorphism of $(\mathbb{Z}/(n))^\times$ to $Z_n$.

1. We saw in a previous theorem that this mapping is surjective; since $Z_n$ is finite, it is bijective. By a previous theorem, $\varphi$ is a homomorphism, thus an automorphism.
2. $(\Rightarrow)$ Suppose $\sigma_a = \sigma_b$. Then $\sigma_a(\alpha) = \sigma_b(\alpha)$, so that $\alpha^a = \alpha^b$. Since $|\alpha| = n$, then, we have $a \equiv b$ mod $n$. $(\Leftarrow)$ Suppose $a \equiv b$ mod $n$; then we have $a-b = kn$ for some integer $k$, so that $a = kn+b$. Now for all $x \in Z_n$ we have $\sigma_a(x) = x^a$ $= x^{kn+b}$ $= x^b$ $= \sigma_b(x)$; hence $\sigma_a = \sigma_b$.
3. Let $\varphi$ be an automorphism of $Z_n$. Now $\varphi(\alpha) = \alpha^k$ for some $k$, since $\alpha$ generates $Z_n$. Then $\varphi(\alpha^i) = \varphi(\alpha)^i$ $= \alpha^{ki}$ $= (\alpha^i)^k$ $= \sigma_k(\alpha^i)$ for all integers $i$; since every element of $Z_n$ is of the form $\alpha^i$, we have that $\varphi = \sigma_k$.
4. Define a mapping $\Phi : (\mathbb{Z}/(n))^\times \rightarrow \mathsf{Aut}\ Z_n$ by $\Phi(\overline{a}) = \sigma_a$. By part 2 above, $\Phi$ is well defined. Moreover, letting $x \in Z_n$ be arbitrary, we have $(\sigma_a \circ \sigma_b)(x) = \sigma_a(\sigma_b(x))$ $= \sigma_a(x^b)$ $= (x^b)^a$ $= x^{ab}$ $= \sigma_{ab}(x)$. Thus $\sigma_a \circ \sigma_b = \sigma_{ab}$. In other words, $\Phi(\overline{a} \overline{b}) = \Phi(\overline{a}) \Phi(\overline{b})$. Thus $\Phi$ is a homomorphism. Finally, by part 3 above, $\phi$ is surjective. Since $\mathsf{Aut}\ Z_n$ and $(\mathbb{Z}/(n))^\times$ are finite $\phi$ is bijective. Thus $\Phi$ is an isomorphism.

### Exhibit the automorphisms of ZZ/(48)

Let $Z_{48} = \langle x \rangle$. For which integers $a$ does the map $\varphi_a$ defined by $\varphi_a(1) = x^a$ extend to an isomorphism $\mathbb{Z}/(48) \rightarrow Z_{48}$?

We know that $\langle x \rangle = \langle x^a \rangle$ precisely when $\mathsf{gcd}(a,n) = 1$. That is, $x^a$ is a generator of $Z_{48}$ precisely when $\mathsf{gcd}(a,n) = 1$. Thus $\varphi_a$ is an isomorphism precisely for 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, and 47.