Find all subgroup of order $8$ of $S_4$ and verify that $N_{2^3} \equiv 1 (\mod 2)$ and divide $3$.

Question

Since $\mid S_4 \mid = 2^3 3$, the permutation group possesses at leat one subgroup of order $8$ (Sylow's theorem). Find all subgroup of order $8$ of $S_4$ and verify that $N_{2^3} \equiv 1 (\mod 2)$ ($N_{p^2}$ is the number of subgroup of cardinality $p^s$) and divide $3$.

I construct a set from the dihedral group $D_4$. I got $$A = \{(1 \ 2 \ 3 \ 4), (1 \ 3)(2 \ 4),\ (1 \ 4 \ 3 \ 2),\ 1_A, (1 \ 3)(2 \ 4),\ (1 \ 4)(2 \ 3),\ (3 \ 4)(1 \ 2)\}.$$

I know how to verify that a set is a subgroup, but this way is very long.

Can you explain me how to show that this set is a subgroup?

Is it the only sub-group of order $8$ of $S_4$?

Answer 1

$S_4$ has $3$ different subgroups of order $8$, which are all isomorphic to the dihedral group $D_4$, see here. We have indeed three different dihedral subgroups in $S_4$, as was explained here. Hence $N_8=3$, which is congruent $1$ modulo $2$. They are subgroups, because they are exactly the Sylow-2-subgroups.