Showing ${\rm Aut}(V) \cong S_3$

Question

I am trying to prove that the automorphisms of the Klein four-group is isomorphic to $S_3$.

The first insight is the Klein group is an abelian group of four elements, three of which are order $2$, and must be mapped to other elements of order $2$, and the identity must of course be mapped to itself.

So any automorphism "permutes" the non-identity elements.

So there are exactly $3! = 6$ of them, matching the number of elements of $S_3$. Then if I call those non-identity maps $f_2, f_2, \ldots, f_6$, I can naturally map them to $S_3$.

For example, if the non-identity elements of the Klein group are $a$, $b$, $c$, I can regard these as $1$, $2$, and $3$ (I believe the labelling doesn't matter) and if I swapped $a$ and $b$ and fixed $c$, I can regard this as the transposition $(12)$.

By repeating this, I get a natural bijection of sets.

I don't know how to "prove" it's a bijection other than to observe the mapping and that it hits everything in $S_3$ exactly once.

I don't know, in particular, how to prove that this is a homomorphism, other than by checking every possible pair of isomorphisms, which is a bit laborious.

I assume there is a better, cleaner way to do it.

UPDATE: After working on this problem more, I have an updated attempt. It's incomplete, but I think it makes some progress.

We realize the Klein 4-group as $V = \{e, a, b, c\}$, where $a,b,c$ are the non-identity elements, each of which has order $2$. Any automorphism $f$ of $V$ must send elements of order $2$ to elements of order $2$, so necessarily it must permute the elements $a,b,c$. I claim that such a permutation is necessarily an automorphism. We observe that the Klein four-group has the property that if we multiply two non-identity elements, we get the third. Since $f$ maps from $V$ to $V$ bijectively, it is also the case that when $f(g)f(g') = f(g'')$, where $g$ and $g'$ are distinct elements and $g''$ is the third. That is, for any $g \neq g' \neq e$, we have \begin{align*} f(gg') = f(g'') = f(g)f(g''). \end{align*} If we take an arbitrary $g \in V$ and then consider multiplying with the identity $e$, we have \begin{align*} f(ge) = f(g) = e f(g) = f(e) f(g), \end{align*} since $f(e) = e$. This proves that $h$ is a homomorphism, and since it is bijective by definition, it is an automorphism. Since every automorphism must send elements of order $2$ to elements of order $2$ and there are exactly $3! = 6$ permutations of $\{a,b,c\}$, there are exactly six elements of Aut(G).

An automorphism is, therefore, uniquely determined on how it permutes the elements ${a,b,c}$, giving a natural bijective map to $S_3$, relabelling these non-identity elements as $1,2,3$. I call $e$ the identity permutation and $f_{abc}$ as the automorphism that maps ${a,b,c} \to {a,b,c}$ (i.e., the identity map) and so forth. The map is: \begin{align*} & f_{a,b,c} \longmapsto e & & \text{both have order $1$} \\ & f_{a,c,b} \longmapsto (23) & & \text{both have order $2$} \\ & f_{b,a,c} \longmapsto (12) & & \text{both have order $2$} \\ & f_{c,b,a} \longmapsto (13) & & \text{both have order $2$} \\ & f_{b,c,a} \longmapsto (123) & & \text{both have order $3$} \\ & f_{c,a,b} \longmapsto (132) & & \text{both have order $3$} \end{align*} This is clearly a bijection of sets: every permutation in $S_3$ was hit exactly once. I still do not know how to show that this is a homomorphism. I constructed the map so that it preserved the order of elements, but is that enough? I could check that $\varphi(xy) = \varphi(x) \varphi(y)$ for each pair, but there must be a way to do it that I'm missing.

Answer 1

Good thing about $V$ is that $V \cong \mathbb{Z}_{2}\times\mathbb{Z}_{2}$. And there is a nice approach, because once you determined where $(1,0)$ and $(0,1)$ go the whole automorphism is defined. So let's represent automorphism as a matrix $$\left(\begin{array}{cc} a & b\\ c& d \end{array}\right),$$ where $(1,0)\mapsto (a,b)$ and $(0,1)\mapsto (c,d)$. It easy to show that such matrix is invertible. More than that, composition of two automorphism is given by multiplication of their matrices. So let $G$ be a group of invertible $2\times 2$ matrices with $\{0,1\}$ coefficients. $|G|=6$ and we can easily list its elements: $$\left(\begin{array}{cc} 1 & 0\\ 0& 1 \end{array}\right),\left(\begin{array}{cc} 1 & 1\\ 0& 1 \end{array}\right),\left(\begin{array}{cc} 1 & 0\\ 1& 1 \end{array}\right),\left(\begin{array}{cc} 1 & 1\\ 1& 0 \end{array}\right), \left(\begin{array}{cc} 0 & 1\\ 1& 1 \end{array}\right),\left(\begin{array}{cc} 0 & 1\\ 1& 0 \end{array}\right) $$ From here we can easily determine "transpositions", because they are of order 2. We can match them with matrices of order 2 and then extend to isomorphism (because transpositions generate symmetric group and such matrices satisfy relations). This is a bit laborious too, but gives some kind of systematic approach.

Answer 2

How this can be proved depends a bit on what knowledge you are allowed to use.

Here is one possible approach, assuming that you have some knowledge on linear algebra over general fields.

The klein group is, in fact, a two dimensional vector space over $\Bbb F_2$. If you compare the definitions of group homomorphisms and $\Bbb F_2$-linear maps, then it should be clear that they are the same thing: the extra requirement of being $\Bbb F_2$-linear is redundant.

Thus the automorphism group of the klein group is simply the group $\operatorname{GL}_2(\Bbb F_2)$ (canonical up to a choice of $\Bbb F_2$-basis). It is a standard exercise in linear algebra that the number of elements of $\operatorname{GL}_n(\Bbb F_p)$ is equal to $\prod_{i = 0}^{n - 1}(p^n - p^i)$, which in our case is $6$.

As the map sending any automorphism of the klein group to the permutation of the three non-identity elements is clearly an injective group homomorphism, by comparing the number of elements we conclude that it is an isomorphism of groups.