Show that $\displaystyle D_4 $ (isometries of a square with vertices 1,2,3,4) is not normal in $\displaystyle S_4$
(Permutation of {1,2,3,4})
$\displaystyle |D_4|=8 $
$\displaystyle |S_4|=4!$
Any help is appreciated! Thanks!
Identify $\displaystyle D_4$ with a definite subgroup of $\displaystyle S_4$ , since you want to show it is not normal there: it will contain two 4-cycles. Now conjugate one of these 4-cycles by a nicely chosen transposition and get a new 4-cycle that isn't contained in your $\displaystyle D_4$ ...
Another way: $\displaystyle D_4$ is a Sylow 2-subgroup of $\displaystyle S_4$ so it is normal there iff there's one unique subgroup of its order (8); Well, show that there are more than one...
Tonio