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I need help to solve this problem: Let $G$ be group of order $231.$ we need to show that the subgroup of order $11$ lies inside $Z(G).$

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You describe "the" subgroup of order 11. Is it clear to you why there is only one (Sylow) subgroup of order 11? – hardmath Feb 6 '13 at 15:53
By this question, it is obvious that there is only one Sylow subgroup of order $11$.So I suppose this is clear to OP? – awllower Feb 6 '13 at 15:54
Sorry, but it appears that my previous arguments used to show that the Sylow subgroup of order $11$ is invalid. One should use instead the theorems of Sylow indeed. Apology here. – awllower Feb 7 '13 at 2:25
Wait, but a more general result than the one in the link could show that $H$ is normal in $G$, without using the theorems of Sylow directly:Since $H$ is a $11$-group, and $|G:H| \equiv -1\pmod{11}$, we can conclude that $|N_GH:H| \equiv -1 \pmod{11}$. But then the only possibility is that $N_GH=G$ and hence $H$ is normal in $G$. – awllower Feb 7 '13 at 4:20
As a possible generalisation, one can consider groups of order $pqr$, where $p$, $q$, $r$ are primes, with $p>q, r$, and where $p-1$ is prime to $q$, and $r$. Then there is only one Sylow subgroup of order $p$, and this subgroup must lie in the center of $G$! Hum, I need to ponder again whether or not the above generalisation is correct... – awllower Feb 7 '13 at 4:37
up vote 9 down vote accepted

Here is an alternative way to do it:

Show that group has a unique subgroup of order $11$ (to make the question make sense). This follows directly from the Sylow theorems.

Let $H$ be this subgroup (which is normal). $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$ by the normalizer/centralizer theorem. But the automorphism group of $H$ has order $10$, which is coprime to $|G|$, so the only possibility is that $C_G(H) = G$, which means that $H$ is central in $G$.

Edit: To see that $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$, we define a map from $G$ to $\rm{Aut}(H)$ by sending each $g\in G$ to conjugation by $g$ (that is, the map from $H$ to itself given by $h\mapsto ghg^{-1}$). Then we check that this map is a homomorphism and has kernel equal to $C_G(H)$.

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Very nice! Action by conjugation $ghg^{-1}$ can be seen to fix everything in $H$ by the coprimality argument, which shows $H$ is in the center of $G$. – hardmath Feb 6 '13 at 16:45
Is this group complete? I mean $G=G'$? – Babak S. Feb 6 '13 at 17:04
@BabakSorouh Such groups are called perfect, and no, any group of order $231$ is solvable (in fact supersolvable). – Tobias Kildetoft Feb 6 '13 at 17:08
+1 Thanks for patience. I see that. – Babak S. Feb 6 '13 at 17:09
@Tobias could you tell me why $G/C_G(H)\cong K\le Aut(H)$ – Un Chien Andalou Feb 6 '13 at 17:14

Here is an outline for a different solution, you can try to fill in the details.

By Sylow's theorem, the group $G$ has a normal subgroup $P$ of order $11$ and also a normal subgroup $H$ of order $7$. Because every group of order $33$ is cyclic, $G/H$ is abelian and $G' \leq H$. Therefore $P$ is a normal subgroup and $P \cap G' = \{1\}$, which implies that $P \leq Z(G)$.

This is same approach sometimes works for similar exercises (that is, exercises like "Prove that subgroup $X$ is central"). However, it does not always work since $P \leq Z(G)$ does not imply $P \cap G' = \{1\}$. In general I would recommend using the normalizer/centralizer theorem as in Tobias answer.

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First, why is every group of order $33$ cyclic? Second, how does one deduce that $P \leq Z(G)$ from the fact that it has trivial intersection with $G'$? Apart from the two points, it is a nice answer. Thanks. – awllower Feb 7 '13 at 2:22
Now I see why all groups of order 33 must be cyclic, so only the second point remains now. – awllower Feb 7 '13 at 4:41
I see all your reasonings now. Thanks for this nice answer. – awllower Feb 7 '13 at 5:05
@awllower: Thanks. Yes, this was meant to be an outline and not a fully detailed answer, perhaps I should have mentioned that. – Mikko Korhonen Feb 7 '13 at 7:10

Yet another approach:
Firstly, since $[G:H] \equiv [N_GH:H] \equiv -1 \pmod{11}$, we conclude that $H$ is normal in $G$. Then, define an action of $G$ on $H$ by conjugation. Again by the class-equation, we have
$|H|=|K| +\Sigma |O_i|$, where K is the set of fixed points and $O_i$ are orbits. But $|H|$ is prime, so either $|K|=0$, or $|K|=11$. Since the identity is always fixed, we conclude that $K=H$, i.e. $H$ lies in the center of $G$.

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Just elementary considerations suffice. – awllower Feb 7 '13 at 4:36
How do you conclude that $[N_G(H):H]$ has the same remainder mod $11$ as $[G:H]$? – Tobias Kildetoft Feb 7 '13 at 14:07
Define an action of $H$ on the set of left cosets of $H$ by conjugation. Then from class-equation we have $[G:H]=|K|+\Sigma_i O_i$, where $K$ is the set of fixed points and $O_i$ are orbits. Then, since $|O_i|$ are all divisible by $11$, we conclude that $[G:H] \equiv |K| \pmod{11}$. But $|K|$ is nothing but $[N_GH:H]$. – awllower Feb 7 '13 at 14:11
And this is contained in my answer to the question that I linked. – awllower Feb 7 '13 at 14:12
Why are the sizes of the orbits divisible by $11$? (the orbits are not the cosets themselves, but sets of cosets). – Tobias Kildetoft Feb 7 '13 at 14:14

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