Proving certain groups are normal Given the following subgroup of $\mathbb S_4$:
$$K=\{\mathrm{id},(12)(34),(13)(24),(14)(23)\}$$
prove that $K \unlhd \mathbb A_4$, $K \unlhd \mathbb S_4$
I am trying to solve this problem doing as little calculations as possible. Given $k \in K$I could try to show $gkg^{-1} \in K$ for any $g \in \mathbb S_4$ but I would like to know if there is a "nicer" solution of the problem. Any ideas or answers would be appreciated. Thanks in advance.
 A: Recall that every permutation has a cycle decomposition as a product of disjoint cycles. This representation determines the cycle type of the permutation as follows: if $\sigma=\tau_1\dots\tau_k$ for each $\tau_i$ a cycle of length $n_i$, with $n_1\geq n_2 \geq n_2 \geq \dots$, then we say that $\sigma$ is of type $n_1,n_2,\dots,n_k$. For instance, in your case all the non-identity elements of $K$ have cycle type $2,2$.
Cycle type is invariant under conjugation (in fact, conjugacy classes are precisely sets of elements of the same cycle type). To see the invariance, take any $\sigma, \mu \in S_n$:
$$\mu \sigma \mu^{-1}(\mu(i))=\mu(\sigma(i)),$$
which shows that the cycle decomposition for $\mu \sigma \mu^{-1}$ is obtained by replacing $i$ with $\mu(i)$ in the cycle decomposition for $\sigma$. That is, if 
$$\sigma = (i_1,\dots,i_l)\cdots (j_1,\dots ,j_m)$$ in cycle notation, then
$$\mu \sigma\mu^{-1}=\big(\mu(i_1), \dots ,\mu(i_l)\big) \cdots \big(\mu(j_1), \dots ,\mu(j_n)\big).$$ 
In particular, $\mu \sigma\mu^{-1}$ is of the same cycle type as $\sigma$. To finish up your problem, just note that your subgroup contains all cycles of type $2,2$.
While this argument requires some calculation, it avoids manipulation of specific permutations. Plus, the general result is good to have on hand.
A: $S_4$ is generated by $(12)$ and $(1234)$ (See comments below. Also take note that if we can get every transposition $(ij)$ then we have all of $S_4$ as every element is the product of transpositions). Show that the Klein-4 group is normal with respect to these two elements. 
For example, let $\alpha = (1234)$. Then $\alpha^{-1}(12)(34)\alpha = (14)(23)$ (Multiplying permutations from right to left). 
Any element of $A_4$ is also in $S_4$. Hence if $K ⊴ S_4$ then $K ⊴ A_4$. 
