# Before proving that $Aut(G)$ is a subgroup of $S_G$ …

Exercise 37 Ch. 9 of the book Abstract Algebra by T Judson :

We will denote the set of all automorphisms of G by $Aut(G)$. Prove that $Aut(G)$ is a subgroup of $S_G$ , the group of permutations of G.

Let $G= \mathbb {Z_6}$. Then there are two isomorphisms between $\mathbb {Z_6}$ and $\mathbb {Z_6}$ with additive operation. One is $\phi (x)=x$ the identity, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{0,1,2,3,4,5}\}$, and the other is $\phi (x)=x+5$, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{5,0,1,2,3,4}\}$. But the set ${\{(1), (054321)}\}$ is not a group at all let alone to be a subgroup of $S_{|\mathbb{Z_6}|}=S_6$.

Where am I wrong?

• $x \mapsto x+5$ is not an additive homomorphism. You probably mean $x \mapsto 5x$. – lhf May 22 '16 at 1:56

The automorphisms of $(\mathbb {Z_6},+)$ are $x \mapsto ax$, for $a=\pm1$.

They form a cyclic group of order $2$, which is a subgroup of $S_6$.

• Is there any homomorphism with $x \mapsto x+a$ ? – Liebe May 22 '16 at 2:02
• @Liebe, just for $a \equiv 0 \bmod 6$ because it has to send $0$ to $0$. – lhf May 22 '16 at 2:07

Your second map is not an automorphism, because it doesn’t even take the identity ($0$) to itself. Rather, the nontrivial automorphism is $x\mapsto-x$, so leaves $0$ and $3$ fixed, otherwise $1\leftrightarrow5$, $2\leftrightarrow4$.