1. ## Subgroup of S4

How can I show that $S_4$ has a subgroup isomorphic to $D_8$? Can someone help please?

2. I wonder who set you this question, because it would seem impossible - indeed it can be shown that no subgroup of $S_{4}$ is isomorphic to $D_{8}.$

Note that $\#D_{8}=16$, eight rotations and eight reflections, and that $\#S_{4}=4!=24.$ By Lagrange's theorem, if $G$ is a subgroup of $S_{4},$ then $\#G|S_{4},$ and it is very clear that isomorphic groups have the same number of elements. But $24$ is not divisible by $16$, so no subgroup of $S_{4}$ is isomorphic to $D_{8}.$

Maybe it was a trick question!

3. Maybe he means $D_4$: some authors write $D_{2n}$ for $D_n$ because that shows the order of the group, while others find the notation $D_n$ more appropriate because of its geometric meaning (and would argue that you write $S_n$ and not $S_{n!}$ !).

In this case it is true, the $2$-sylow subgroups of $S_4$ are isomorphic to $D_4$ (or $D_8$ if its order is $8$).

Think of the symmetries of a square whose vertices are nammed $1,2,3,4$ , take a close look at the subgroup generated by $\{(1234), (12)(34)\}$ for instance.