In Chapter 9 Exercise I4 in Pinter's "A Book of Abstract Algebra", one is asked to prove that the set ${\rm Aut}(G)$ is a subgroup of $S_G$.
While I have successfully constructed a proof that demonstrates closure and inverses (i.e. proving that ${\rm Aut}(G)$ is in fact a group), I am still trying to wrap my head around why exactly ${\rm Aut}(G)$ is not actually equal to $S_G$. From some of the other exercises, I had developed the impression that all permutations (say of the form $\pi_n: G \rightarrow G$) were, in fact, automorphisms.
Could someone please provide me with a general example of $f \in S_G$ but $f \notin {\rm Aut}(G)$?
Cheers~
Best Answer
$\text{Aut}(G)$ is the set of all group automorphisms, that is, bijections which preserve the group structure. $S_G$, however, is the set of all bijections, even those which are not group homomorphisms.
As an explicit example, take any permutation which doesn't fix the group identity. This is a permutation in $S_G$, but cannot be an automorphism.
I hope this helps ^_^