I need some help with this, because I have been struggling on it although it seems really easy.
I wanna show that a group of order $440$ has a unique subgroup of order $55$. It is very easy to see the existence of this subgroup (using the Second isomorphism theorem), but I can't find the way too show that it's unique.
Any help will be appreciated,
Thank you!
Applying Sylow's theorem to the 11-Sylows in $G$ of order 440, we see that $G$ has a unique 11-Sylow, which is thus normal--call it $H$. Now, the subgroups of order 55 in $G$ correspond 1-1 to the subgroups of order 5 in $G/H$ which are their images. (If this isn't obvious at first, note that since $H$ is the only order 11 subgroup of $G$, it must be a subgroup of every order 55 subgroup of $G$, by Sylow applied to the order 55 subgroup.) But $G/H$ has order 40 and clearly (by Sylow) has only 1 subgroup of order 5. Thus, $G$ has only one subgroup of order 55.