# Show that a group of order 440 has a unique subgroup of order 55

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.
• Yes, you can have 11 5-Sylows in $G$, but they would all be subgroups of the unique subgroup of order 55. – C Monsour Oct 23 '18 at 0:23