Why any permutation defines an automorphism?

Question

My professor kept on saying that: any permutation defines an automorphism.

My questions are:

1-Is there a rigor proof for this fact please?

I know that a permutation of a set $M$ is a bijective function from M to itself. And I know that an automorphism is an isomorphism from M to itself and that makes every permutation an automorphism. Is this the only reason for saying that any permutation defines an automorphism.

2-And what is the normal automorphism defined by any permutation?

Answer 1

An automorphism of a structure $\mathfrak{A}$ - being deliberately loose about what I mean by "structure" - is a permutation of the elements of $\mathfrak{A}$ which preserves all the relevant operations and relations. For example, if $\mathfrak{A}$ is just a set with no additional structure then "automorphism of $\mathfrak{A}$" is the same as "permutation of $\mathfrak{A}$," and this is what Qiaochu's initial comments above were getting at. By contrast, an automorphism of a group is more complicated: it has to preserve the "group structure," which is to say it has to satisfy the following rules:

• $f(e)=e$,

• $f(x^{-1})=f(x)^{-1}$, and

• $f(x*y)=f(x)*f(y)$.

Any nontrivial group has permutations which are not automorphisms (just consider any permutation moving the identity). So taken too literally, the slogan "Any permutation defines an automorphism" is blatantly false for every interesting group.

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However, that's not what your professor meant: there's a more nuanced interpretation which some groups (such as the Klein four group $V$) satisfy while others don't (exercise).

Specifically, some structures - and in particular, some groups - have a very nice property: that their automorphisms "are" permutations of some other set. This is especially interesting when that other set is a subset of the structure itself.

The Klein four group $V$ provides an example of this. Let $a_1,a_2,a_3$ be the non-identity elements of $V$; every permutation $\pi$ of $\{a_1,a_2,a_3\}$ extends to a unique automorphism $\alpha_\pi$ of $V$, and conversely every automorphism of $V$ restricts to a permutation of $\{a_1,a_2,a_3\}$.

This is not in general the case! If $G$ is an arbitrary group, there's no reason for $Aut(G)$ to be isomorphic to $S_X$ for any $X$. So we're seeing here a very special property of $V$.

And as a follow-up exercise:

Suppose $G$ is a finite group, $n\in\mathbb{N}$, and $Aut(G)=S_n$. Need there be some $\{a_1,...,a_n\}\subseteq G$ such that every permutation of $\{a_1,...,a_n\}$ extends to a unique automorphism of $G$?