Coming from a non-group theory background, I noticed that the finite groups I was dealing with seem to all have the following property. Let $G$ be a finite group, $H$ a subgroup. Then the normalizer $N\_G(N\_G(H))$ of the normalizer of $H$ is just $N\_G(H)$. It seems to be an exercise in almost any group theory book that this is true if $H$ is a Sylow subgroup, and it is definitely not true in general. The easiest examples are the dihedral group $D\_8$of order 8, and similar examples for groups of order $p^3$, for any odd prime $p$. But these groups have an element of order $p^2$. Is there an easy example where the $N\_G(H) = N\_G(N\_G(H))$ does not hold for some subgroup $H$ of $G$, where $G$ has no elements of order $p^2$, for any prime $p$ (including 2)? 



Normalizer of a Normalizer of a subgroup of a finite group with no elements of order $p^2$
No, take the wreath product $G$ of $C\_3$ and $C\_2$, two cyclic groups of orders 3 and 2. It has order $18$, and an abelian normal subgroup $A=C\_3\times C\_3$. Let $H$ be one of the factors $C\_3$ in $A$. Then $N(H)=A, N(N(H))=G$ and $G$ does not have elements of order $p^2$ for any $p>1$. 


