Looking for a more elegant proof that $H$, a subgroup of $A_4$, is isomorphic to the Klein $4$-group

Question

Is there a better, more general way to define an isomorphism between the Klein $4$-group, $V$, and a subgroup of $A_4$, the alternating group?

First, my definition of $V$ is $$\{\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix}-1 & 0\\ 0 & 1\end{bmatrix},\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix},\begin{bmatrix}-1 & 0\\ 0 & -1\end{bmatrix}\}$$ Currently, I am defining a subgroup $H$ of $A_4$ where $H=\{(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$. Note: the elements of $H$ are written in cycle form simply for my convenience; $()$ is my notation for the identity element in $A_4$.

First, I prove $H$ is a subgroup. I have no problem with this, though my proof is a bit clunky.

Then, I define my isomorphism $\phi:H\rightarrow V$ explicitly. That is, I write out each element of $H$ and what it maps to in $V$. I know that this will $eventually$ work, but I cannot help but wonder if there is a more intelligent way to define my isomorphism which will save time.

Answer 1

Since the Klein 4 group $V$ is the only group of order 4 with 3 elements of order 2 (up to isomorphism) your subgroup $H$ has to be isomorphic to $V$.

Answer 2

Any group of order 4 in which every element has order 2 has the same multiplication table: $$ \begin{array}{c|cccc} \cdot&e&a&b&c\\\hline e&e&a&b&c\\ a&a&e&c&b\\ b&b&c&e&a\\ c&c&b&a&e \end{array} $$ Hence, they are isomorphic.