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There is a hint:

In the case that $G$ does not have a normal subgroup of order 11, prove that for an element $x\in G$ of order not 1 or 11, the centralizer $C_G(x)$ of $x$ has order 12.

I try $N/C$ theorem, but it does not work.

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  • $\begingroup$ If the $11$-subgroup is not normal, there are more than one $11$-subgroups. $\endgroup$ – ajotatxe Dec 15 '17 at 22:24
  • $\begingroup$ Something about this problem is a bit odd. I can't picture a group of order $12\cdot11$ that has a normal subgroup of order twelve but does not also have a normal subgroup of order eleven also! Otherwise the group would need to be a semidirect product. But no group of order twelve has an automorphism of order eleven??? May be this is to be a step in a longer argument leading to the conclusion that such a group always has a normal subgroup of order eleven??? $\endgroup$ – Jyrki Lahtonen Dec 16 '17 at 9:18
  • $\begingroup$ This question is misleading because $|G|=11\cdot 12$, and, $12$ not being a prime power, there must be a normal subgroup of order $11$. However, I think this is pretty typical of introductory Sylow-type exercises in general. $\endgroup$ – Badam Baplan Dec 21 '17 at 20:25
  • $\begingroup$ For example, a standard exercise is to show that the group of order $132$ (or some other nice number) isn't simple, and a standard proof assumes that every $p$-subgroup occurs more than once and reaches a contradiction by a counting argument. Similar to the phrasing of this question, these proofs are (often) inherently weak: a conjunction of premises is shown to be false when in fact some of them individually are necessarily false. But I wonder what the intent of the author of this question was, perhaps just an oversight $\endgroup$ – Badam Baplan Dec 21 '17 at 20:25
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As Jyrki Lahtonen points out in the comments, this question is a bit odd, misleading actually.

It makes more sense to approach the problem in slight generality.

Proposition: Let $p$ be a prime and $G$ be a group of order $|G| = p(p+1)$. Then $G$ has a normal subgroup of order $p$ or a normal subgroup of order $p+1$. If $p+1$ is a not a prime power, then $G$ necessarily has a normal subgroup of order $p$.

First we note that this is certainly true for the first case, $|G|=2\cdot3 = 6$. There's one group of order $6$ and it has normal subgroups of order $2$ and order $3$.

In the general case.... from Sylow theorems, immediately either (1) $G$ has a normal subgroup of order $p$ (nothing to prove) or (2) the number $n_p$ of $p$-subgroups in $G$ is $p+1$. So suppose the latter. Since $p$-subgroups have prime order, they are cyclic, and their intersections are trivial. It follows that there are exactly $(p+1)(p-1)$ elements of order $p$ in $G$, and exactly $p$ non-identity elements of order other than $p$. Let $X$ be the set of all $p$ non-identity elements of order other than $p$.

Pick any $g \in X$ and consider the action of any $p$-subgroup $P$ on $g$ by conjugation. Since $P$ is cyclic, we deduce two possibilities: (a) the action fixes $g$ (b) the action generates $X$. However, recall that from Sylow theorems we have $p+1 = n_p = \frac{|G|}{|N_G(P)|}$, from which it follows that $|N_G(P)| = p$ and $P$ is self-normalizing. We see that (a), which would imply that $g \in N_G(P)$, is actually impossible. Hence (b) holds for every $p$-subgroup, and $g$ is conjugate to every element of order not equal to $1$ or $p$. Since conjugation preserves order, we see that every such element has the same order.
In particular, Cauchy's theorem (any prime divisor of $G$ is order of some element) shows that $G$ can have at most one prime divisor beyond $p$. We have thus shown that if $G$ does not have a normal subgroup of order $p$, then $p+1$ must be a prime power. Contrapositively, a group of order $p(p+1)$ with $p+1$ not a prime power must have a normal subgroup of order $p$.

The last thing to check is that if $p+1 = q^n$ for $q$ prime and $G$ has no normal subgroup of order $p$, then it has a normal subgroup of order $p+1$. Since from the preceding work there are exactly $q^n$ elements available for $q$-subgroups, we see by Sylow that there is precisely one $q$ subgroup (of order $q^n = p+1$), and it is normal. $\Box$

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  • $\begingroup$ We have one or four 3-Sylow group(s). Since a Sylow p-group has empty intersection with center, either choice will take out at least two candidates from the centralizer of $x$, so the question is wrong? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 23:09
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    $\begingroup$ I am clear about what you said, but I want to know, in your way, how to prove the rest 12 elements constitute a normal subgroup? $\endgroup$ – HeHe Dec 15 '17 at 23:13
  • $\begingroup$ Chiming in with HeHe's comment. This is also how I started. But why do those twelve elements form a subgroup? If they do that subgroup is necessarily normal, but why? What is the relation to the hint? No fault with what you have shown, Badam. But I don't see a road to the destination yet :-) $\endgroup$ – Jyrki Lahtonen Dec 16 '17 at 9:12
  • $\begingroup$ OK I added some more details! hope that clarifies everything, this was a funny question. I don't go the route of proving that the rest of the 12 elements constitute a normal subgroup, because that is misleading in itself, since the premises leading to that scenario are impossible. $\endgroup$ – Badam Baplan Dec 21 '17 at 20:24
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Do you know Sylow's Theorem? If so, apply it to the subgroups of order 11.

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  • $\begingroup$ a group of order 11 is always cyclic, and then? $\endgroup$ – HeHe Dec 15 '17 at 22:37

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