1
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ $x \mapsto x+5$ is not an additive homomorphism. You probably mean $x \mapsto 5x$. $\endgroup$ – lhf May 22 '16 at 1:56
2
$\begingroup$

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 $S_6$.

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

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 $1\leftrightarrow5$, $2\leftrightarrow4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.