Intersection of Center of p-group and nontrivial normal subgroup is nontrivial

Let G be a p-group and H a nontrivial normal subgroup. Prove that H\cap Z(G) is nontrivial.

Let G act on H by conjugation. Since H is a normal subgroup, this is a well-defined group action since ghg^{-1}\in H for all g\in G.

Let H_0=\{h\in H:ghg^{-1}=h\ \ \ \forall g\in G\}.

h\in H_0 \iff O(h)=\{h\}.

Therefore we have H=H_0+O(h_1)+O(h_2)+\dots+O(h_n) where |O(h_i)|>1.

By the Orbit-Stabilizer Theorem,

|H|=|H_0|+\sum_{i=1}^k |O(h_i)|=|H_0|+\sum_{i=1}^k[G:G_{h_i}]

Let |G|=p^n. By Lagrange’s Theorem, p^n=|G|=[G:G_{h_i}]|G_{h_i}|. Since |G_{h_i}|\neq p^n, therefore p\mid [G:G_{h_i}].

Hence, |H|\equiv |H_0|\pmod p.

Note that

\begin{aligned} h\in H_0\subseteq H &\iff ghg^{-1}=h\ \ \ \forall g\in G\\ &\iff gh=hg\ \ \ \forall g\in G\\ &\iff h\in Z(G)\cap H \end{aligned}

Therefore |H\cap Z(G)|\equiv |H|\equiv 0\pmod p. Since 1\in H\cap Z(G), this implies that p \mid |H\cap Z(G)| \neq 0. Therefore |H\cap Z(G)|\geq p.

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Author: mathtuition88

Math and Education Blog

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