# Let $|G|=2^np$, $p$ an odd prime, $H\unlhd G$ a Sylow $2$-subgroup with $H\cong(\Bbb{Z}/2\Bbb{Z})^n$, $p\nmid 2^n-1$. Prove $Z(G)$ is nontrivial.

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?

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 $$p$$.
• 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. Jun 4, 2020 at 2:35