Show that the Sylow 11-subgroup of a group of order 990 is normal The SmallGroups library in the program GAP reveals that each of the thirty groups of order 990 has a normal subgroup of order 11. But I would like to prove this fact directly, without reliance on a computer. The alternative to a single subgroup of order 11 is that there are 45 of them but I see no immediate way to rule that out....
 A: Every finite group $G$ of order $2m$, where $m$ is odd, has a subgroup of order $m$ which, since it has index $2$ in $G$, is normal in $G$ (see this for instance). Thus a group of order $990$ has a normal subgroup $H$ of order $495=3^2 \cdot 5 \cdot 11$. Now focus on $H$. (Note that $n_3(H) \in \{1, 55\}$, $n_5(H) \in \{1, 11\}$ and $n_{11}(H) \in \{1, 45\}$.)
A: Following on the hints from the first answer provided (thanks!) and attempting to work through this in my own words... here is my argument:
(1) Left multiplication provides an action of the group $G$ on itself and in this action, each nonidentity element is a permutation (in $S_{990}$) which fixes no points.
(2) An element of order two exists (Cauchy/Sylow) and such an element must be a product of 495 transpositions, thus an odd permutation.
(3) The intersection of the image of $G$ in $S_{990}$ thus intersects $A_{990}$ in a proper subgroup $H$ of order 495.
(4) If the Sylow 11-subgroup of $H$ is not normal then it has 45 conjugates, giving $45\cdot 10 = 450$ elements of order 11. This forces a unique Sylow 5-subgroup $K$.
(5) $H/K$, a group of order 99, has a unique Sylow 11-subgroup $P/K$. 
(6) Any Sylow 11-subgroup of $H$ must be mapped onto $P/K$ by the natural homomorphism and it is not possible for the 5-1 natural map to send all 45 subgroups of $H$ to $P/K.$
The contradiction in (6) shows that the assumption in (4) is invalid and so the Sylow 11-subgroup of $H$ is unique.  Since any Sylow 11-subgroup of $G$ is a subgroup of $H$, then $G$ has only one Sylow 11-subgroup.
