Considering
$K = \{ (1); (1 2)(3 4); (1 3)(2 4); (1 4)(2 3) \} \subseteq S_4$
a) Prove $K$ is a subgroup of $A_4$ and $S_4$
b) Prove that for every transposition $\tau \in S_4$, $\tau K \tau = K$
C) Prove that K is a normal subgroup of S4 and A4
So I have already done a) and b). And for c) the given solution of the exercise states that because $A_4$ is normal in $S_4$, it is enough to verify it for $A_4$. I have verified it for $A_4$ using the result in b).
What I don't get is why they say that "because $A_4$ is normal in $S_4$, it is enough to verify it for $A_4$"? how does normality of $A_4$ imply that $K$ is also normal in $S_4$?. I read somewhere that it is not true that a normal subgroup of a normal subgroup is also normal in the group, so that couldn't be it. Any help is welcome.
Best Answer
Since each $\sigma\in S_4$ can be written as $\sigma=\tau_1\circ\ldots\circ\tau_{n}=:\tau_1\tau_2\ldots\tau_n$ ($\tau_i$ is a transposition for $i=1,\ldots,n$), you have $$\sigma K\sigma^{-1}=(\tau_1 \ldots \tau_n) K(\tau_1\ldots\tau_n)^{-1}=\tau _1 \ldots\tau_{n-1} \tau_nK\tau_{n}^{-1}\tau_{n-1}^{-1}\ldots\tau_1$$ Now use $\tau_i K\tau_i^{-1}=K$ for $i=1,\ldots, n$ to arrive at $\sigma K\sigma^{-1}=K$. Since $\sigma$ is arbitrary, $\sigma K\sigma^{-1}=K$ for all $\sigma \in S_4$