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Let $V$ denote the Klein four-group. I am trying to show that $\mathrm{Aut}(V) \cong S_3$. For this moment, I have a very specific question on the proof.

A key part of the proof is showing that $|\mathrm{Aut}(V)| = 6$. Every automorphism of $V$ needs to send the identity to the identity. $V$ then has three elements $a,b,c$ of order $2$, with the property that the product of any two gives the three. An automorphism needs to send an element of order $2$ to another element of order $2$, thus "permuting" the elements $a,b,c$. One such automorphism, for example, is the identity automorphism.

Any automorphism has to have this form because it has to preserve the order of elements, but I don't know how to check that such a map is in fact an automorphism. That it's a bijection is clear from the map itself: I just need to show that it preserves multiplication in $\mathrm{Aut}(V)$ .

I'm going to try to show this for one proposed automorphism. I'll denote by $f_{a,c,b}$ the automorphism that sends $$ e \mapsto e, a \mapsto c, c \mapsto b, b \mapsto a. $$ I need to show that for all $x,y \in V$, $f_{a,b,c} (x,y) = f_{a,b,c} (x) f_{a,b,c} (y)$. One way to do this is surely by brute force, taking all possibilities for $x$ and $y$. That would mean $4^2 = 16$ possibilities. But I think there must be something I'm missing that would allow me to generalize this argument to any such automorphism, say $f_{\mathrm{Perm}(\{a,b,c\})}$.

Any advice on how to show this would be appreciated. After checking that $\mathrm{Aut}(V)$ has $6$ elements, I believe the most efficient way to show it is isomorphic to $S_3$ is to show that it is non-abelian, and the only non-abelian group of order $6$, up to isomorphism, is $S_3$.

Shaun
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Mathematical Endeavors
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  • Compare with the common proofs, e.g., here. – Dietrich Burde Mar 07 '23 at 22:59
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    By "the" proof, you mean a proof . . . – Shaun Mar 07 '23 at 23:23
  • No, you do not need to do all of them. It is clear that $f_{abc}(xy)=f_{abc}(x)f_{abc}(y)$ when one of them is $e$ and when $x=y$, and because the group is abelian, checking it for $xy$ is the same as checking it for $yx$. That means that you really only need to check three products: $ab$, $ac$, and $bc$. – Arturo Magidin Mar 08 '23 at 01:57
  • @ArturoMagidin I considered the symmetry in $V$, because $xy = yx$, but $S_3$ isn't abelian, so I'm not certain that $f_{abc} (x) f_{abc} (y) = f_{abc} (y) f_{abc} (x)$. Am I missing something? – Mathematical Endeavors Mar 08 '23 at 04:57
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    The values of $f_{abc}$ are in the Klein 4-group, not in $S_3$. You are checking that $f_{abc}$ is an automorphism of $V$, not mapping the automorphism group of $V$ to $S_3$, so $S_3$ isn't even in the room, so the fact that it is not abelian is irrelevant. – Arturo Magidin Mar 08 '23 at 05:07

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