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.

  • $\begingroup$ But there can be either 1 or 11 5-Sylow's, isn't that right? $\endgroup$ – Phil Mett Oct 23 '18 at 0:21
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    $\begingroup$ Yes, you can have 11 5-Sylows in $G$, but they would all be subgroups of the unique subgroup of order 55. $\endgroup$ – C Monsour Oct 23 '18 at 0:23
  • $\begingroup$ Okay thanks. About this: "G/H has order 40 and clearly (by Sylow) has only 1 subgroup of order 5.". Is this true because all 5-Sylows must be disjoint? $\endgroup$ – Phil Mett Oct 23 '18 at 0:26
  • $\begingroup$ Right, because none of 2, 4, or 8 are congruent to 1 mod 5 $\endgroup$ – C Monsour Oct 23 '18 at 0:26

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