How to show that Klein four-group is a normal subgroup of the alternating group $A_4$

I want to show that the Klein four-group is a normal subgroup of the alternating group $A_4$.

I am using the information in this link, that shows explicitly $A_4$, and Klein four-group as a subgroup.

I know that there is the direct way, by definition, but is there a way that does not require actually multiplying so many permutations ?

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The elements of the Klein $4$-group sitting inside $A_4$ are precisely the identity, and all elements of $A_4$ of the form $(ij)(k\ell)$ (the product of two disjoint transpositions).

Since conjugation in $S_n$ (and therefore in $A_n$) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

You do not have to compute the entire Cayley table of a group to get an isomorphism. I think that this is what you are struggling with so I’ll put a little of detail.

First of all, there are only two groups of order $4$ : the cyclic group of order $4$ and the Klein group, it is quite easy to see it. Let $G$ be a group of order $4$ : by Cauchy’s theorem, since $2$ divides $4$ and $2$ is prime, there exists an element of order $2$ in $G$. Let $a$ be this element and $H = \langle a \rangle$. Since $[G:H] = 2$, $H \triangleleft G$, hence for all $g \in G$, $gag^{-1} \in H$, but since $a \neq 1$, we must have $gag^{-1} = a$, that is, $a$ commutes with every element of $G$.

Consider an element outside $H$, call it $b$. If it has order $4$, then $G = \langle b \rangle \cong C_4$, the cyclic group of order $4$. If not, then since the order of an element divides the order of the group, and $b \neq 1$, then $b$ must have order $2$. But then the fourth element cannot have order neither $1$ or $4$, hence it must have order $2$ too. Hence every non-trivial element of $G$ has order two, and using the argument above, they commute with each other ; this gives you the group $C_2 \times C_2$, which is precisely the Klein group, because we must have $ab = c$, $ac = b$ and $bc = a$ (for obvious reasons, because other possibilities lead to contradictions).

Now the subgroup of $A_4$, namely $K=\{(1), (12)(34),(13)(24),(14)(23) \}$ is a subgroup of order $4$. Since it is not cyclic, it is isomorphic to the Klein group. Conjugation in $S_n$ does not change cycle structure, so that in particular it does not do that in $A_n$. This means that this subgroup is normal, because $gKg^{-1} \subseteq K$, which is an equivalent condition for normality of a subgroup.

Hope that helps,

P.S. : Maybe I used sledgehammers to classify groups of order $4$, but I just threw out the first ideas that came to mind ; if you want to simplify them by commenting feel free.