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 ?
 A: 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.
A: 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.
A: You can do this without resorting to conjugacy classes or the result on cycle types. It is elementary that the conjugates of any element have the same order as the original element.
The non-identity elements of the identified Klein 4 subgroup are the only elements of $A_4$ which have order 2 (the other eight elements have order 3). The only possible images under conjugation therefore lie within the group. It is easy to complete this observation to a proof.
