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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....

Shaun
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KenWSmith
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  • All groups of order 990 also have a normal Sylow 3-subgroup. Proving that by hand also seems rather difficult to me. – KenWSmith Nov 25 '19 at 13:23
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    If you show that $n_3=1$ first, I think that the result follows easily because the groups of order $3 \cdot 11$ are cyclic. – Adler Marques Nov 25 '19 at 13:40

2 Answers2

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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 $n_{11}(H) \in \{1, 45\}$.)

the_fox
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    Let me know if you want me to finish the argument. You just have to play around a little with the possibilities. – the_fox Nov 25 '19 at 15:09
  • Thanks! I see two parts to work through -- the first is the claim on $2m$, $m$ odd (thanks for the link) and the second is the claim about $n_{11}(H).$ I am pretty sure I can do the second, since I had already spent a lot of time working through $N(P)$ of order 22, etc., without the advantage of this big normal subgroup. (I now recall seeing the $2m$ claim elsewhere, but not in a standard textbook environment.) – KenWSmith Nov 26 '19 at 00:30
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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.

KenWSmith
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  • This argument, based on the existence of $H$, also shows why the Sylow 3-subgroup is unique. By removing the factor 2 in 990, a lot of extraneous choices collapse. – KenWSmith Nov 29 '19 at 17:15