Seeking a general example of $f \in S_G$ but $f \notin {\rm Aut}(G)$

Question

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~

Answer 1

$\text{Aut}(G)$ 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.

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I hope this helps ^_^

Answer 2

Hint If $G$ is not commutative, the mapping $x \to x^{-1}$ is such an example.

Answer 3

Let’s take a very easy example. Let $G = \mathbb{Z}_2.$ Then $S_G$ is precisely $\{f_1, f_2\}$ where $f_1$ is the identity map and $f_2$ is simply the map which “switches” the only two elements of $\mathbb{Z}$. That is, $f_2(0) = 1$ and $f_2(1) = 0.$

Now, observe that $f_2$ is a bijection which doesn’t fix the identity element. This implies that $f_2$ cannot be a group homomorphism and hence is not an automorphism of $\mathbb{Z}_2.$ So, $\text{Aut}(G) = \{f_1\} \neq S_G$ in this case.

Answer 4

Write down a bijection that maps an element x -> y where $|x|$ $\neq$ $|y|$. This could be written as (xy) in permutation cycle notation.

This works because automorphisms are isomorphisms which must preserve order of elements.

Answer 5

Left multiplication by a given nontrivial element $a\in G$ is a bijection on $G$ (this follows from group axioms), but it isn't an endomorphism, as: $$agh=agah\iff a=e$$