Prove that a group of order $132=11\cdot 12$ has a normal subgroup of order 11 or a normal subgroup of order 12. 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. 
 A: 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.  $\Box$
A: Do you know Sylow's Theorem? If so, apply it to the subgroups of order 11.
