TITLE: Is it possible for a group of order 132 to have 12 different subgroups of order 11?
QUESTION [5 upvotes]: For me it looks like it isn't possible, but I can't see an argument. I've tried something with Sylow groups (consider all of them as Sylow 11-group and try to use Sylow theorems to find a contradiction), but unfortunately it didn't bring much. 
Thanks in advance!

REPLY [10 votes]: I’m not any kind of group theorist, so I hope I don’t get egg on my face here. Look at those $12$ groups of order $11$. Any two of them can intersect only in the identity (otherwise they would be the same), so you have $12\times10$ elements of order $11$ in your group. This leaves only $12$ more: the identity and some elements of order $3$, $2$, and possibly $4$. Take any element $g$ of order $11$, and use it to conjugate and element $h$ of order $3$. If $ghg^{-1}$ is not $h$, then the elements $g^nhg^{-n}$ are eleven in number, eleven different elements of order $3$, leaving no room for elements of order $2\,$! So $g$ and $h$ must commute. Same argument for elements of order $2$ or $4$, so any element of order $11$ commutes with an element of order prime to $11$. This means the group is commutative after all, and then there is only the one subgroup of order $11$.
(I will defer to anyone that gives a less messy argumenent!)