Let $G$ be a group with $|G|=2^np$ ($p$ an odd prime). Let $H$ be a normal Sylow $2$-subgroup such that $H\cong(\mathbb{Z}/2\mathbb{Z})^n$. Prove that if $p$ does not divide $2^n-1$, then $G$ has a non-trivial center.

I think this may have something to do with the fact that $H$ can be written as a union of conjugacy classes and then apply the class equation, but I have no idea. Furthermore, we probably have to make use of the fact that since $H$ is normal it is the unique Sylow $2$-subgroup of $G$. Any hints?


1 Answer 1


Suppose $G$ has trivial center. Let $x\in H$. Since $H$ is abelian, the centralizer $C_G(x)$ contains $H$, so either $C_G(x)=H$, or $C_G(x)=G$. However, $C_G(x)=G$ if and only if $x$ is in the center of $G$ if and only if $x$ is the identity. Thus, $C_G(x)=H$ for all $x\in H$ except the identity. Hence, $\vert\text{Cl}(x)\vert=[G:C_G(x)]=p$ for every $x\in H$ except the identity. Because $H$ is normal, $\text{Cl}(x)\subset H$ for every $x\in H$. Thus, $H$ is the disjoint union of sets of size $p$ and the trivial group, so $\vert H\vert-1$ is divisible by $p$.

  • $\begingroup$ You should use $C_G(x)$, the centralizer of $x$. Less risk of confusion than the normalizer, which is usually applied to subgroups, not subsets. $\endgroup$ Jun 4, 2020 at 2:35
  • $\begingroup$ @ArturoMagidin Sure, if you say so. $\endgroup$
    – Anonymous
    Jun 4, 2020 at 2:39

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