How do I show that the subgroup $H_2=\{\text{id}, (1\quad 2)(3\quad 4),(1\quad 3)(2\quad 4),(1\quad 4)(2\quad 3)\}$ is normal in $S_4$?
The tip that I got was to use the fact that for every $\tau\in S_4$ and every k-cycle $\sigma=(x_1\quad x_2\quad...\quad x_k)\in S_4$ we know that $\tau\sigma\tau^{-1}=(\tau(x_1)\quad\tau(x_2)\quad...\quad\tau(x_k))$. But I don't see how.
For showing $H_2$ is normal you need to show that $\forall \tau \in S_4$ we have $\tau H_2 \tau ^{-1} = H_2$ (i.e. invariant under conjugation).
Let $\alpha \in H_2$ and $\alpha \neq (1)$. Then $\alpha$ is a $(2,2)$ cycle, i.e $\alpha=(a \, b) (c \, d)$. Then \begin{align*} \tau \alpha \tau^{-1} & = \tau (a \, b) (c \, d) \tau^{-1}\\ &=\tau (a \, b) \color{red}{\tau^{-1} \, \tau}(c \, d) \tau^{-1}\\ &=(\tau(a) \, \tau(b)) \, (\tau(c) \, \tau(d)). \end{align*} Since $\tau$ is a permutation, therefore $(\tau(a) \, \tau(b)) \, (\tau(c) \, \tau(d))$ is also a $(2,2)$ cycle. But all $(2,2)$ cycles are in $H_2$, therefore $H_2$ is invariant under conjugation, hence normal.
Note: For sake of completion one should also observe that $\tau (1) \tau^{-1}=(1) \in H_2$.
