Normal subgroup in S4

Question

Let H be a subgroup of S4 where $H = \{e, B , C ,D \}$

$B(1)=2,B(2)=1,B(3)=4,B(4)=3$

$C(1)=3,C(2)=4,C(3)=1,C(4)=2$

$D(1)=4,D(2)=3,D(3)=2,D(4)=1$

Prove that H is a normal subgroup.

I've tried using the definition that $H$ is normal $\iff gH=Hg$, $\forall g$

As I can't take all the 24 elements of S4 and check it,I've tried with a general form: $g(1)=a,g(2)=b,g(3)=c,g(4)=d$. I managed to find $gH$,but $Hg$ isn't as easy.

What I've also noticed is that $BC=D,CD=B,DB=C, B=B^{-1} , C=C^{-1} , D=D^{-1}$ and $B^{2}=C^{2}=D^{2}=e$.

Can you give me a hand, please?

Answer 1

Clearly this subgroup is a subgroup of the alternating group. Notice that any isomorphism from $A_{4}$ to $A_{4}$, must carry elements of order 2 to elements of order 2 and since $A_{4}$ contains 3 elements of order 2, e, and 8 elements of order 3,any isomorphism from $A_{4}$ to $A_{4}$ must carry $H$ to $H$. Now $A_{4}$ is a normal subgroup of $S_{4}$, and hence conjugation by elements of $S_{4}$ defines an isomorphism from $A_{4}$ to $A_{4}$. And hence must fix $H$, so $gHg^{-1}=H$ for all $g \in S_{4}$.

Answer 2

For any $\sigma\in S_4$ the conjugate of $B$ will have the same cycle structure as $B$, and similarly for $C$ and $B$, since $$\sigma(i_1\dots i_r)\sigma^{-1}=(i_{\sigma(1)}\dots i_{\sigma(r)}) $$ Now these are the only permutations in $S_4$ with a $(2,2)$ structure. Hence the conjugates of each of $B, C, D$ is one of them. As for $e$, its conjugates are $e$ itself. Hence $H$ is normal.