How can I show that $\displaystyle S_4$ has a subgroup isomorphic to $\displaystyle D_8$? Can someone help please?
I wonder who set you this question, because it would seem impossible - indeed it can be shown that no subgroup of $\displaystyle S_{4}$ is isomorphic to $\displaystyle D_{8}.$
Note that $\displaystyle \#D_{8}=16$, eight rotations and eight reflections, and that $\displaystyle \#S_{4}=4!=24.$ By Lagrange's theorem, if $\displaystyle G$ is a subgroup of $\displaystyle S_{4},$ then $\displaystyle \#G|S_{4},$ and it is very clear that isomorphic groups have the same number of elements. But $\displaystyle 24$ is not divisible by $\displaystyle 16$, so no subgroup of $\displaystyle S_{4}$ is isomorphic to $\displaystyle D_{8}.$
Maybe it was a trick question!
Maybe he means $\displaystyle D_4$: some authors write $\displaystyle D_{2n}$ for $\displaystyle D_n$ because that shows the order of the group, while others find the notation $\displaystyle D_n$ more appropriate because of its geometric meaning (and would argue that you write $\displaystyle S_n$ and not $\displaystyle S_{n!}$ !).
In this case it is true, the $\displaystyle 2$-sylow subgroups of $\displaystyle S_4$ are isomorphic to $\displaystyle D_4$ (or $\displaystyle D_8$ if its order is $\displaystyle 8$).
Think of the symmetries of a square whose vertices are nammed $\displaystyle 1,2,3,4$ , take a close look at the subgroup generated by $\displaystyle \{(1234), (12)(34)\}$ for instance.