Number of inequivalent one-dimensional representations of a group of order $p^n$

Question

This is from an old comprehensive exam:

Prove that any group with $p^n$ elements has a nontrivial center and use this to prove that any group with $p^n$ elements has at least $p$ inequivalent one-dimensional complex representations. Here $p$ is a prime number.

The part about having a nontrivial center is easily done with the class equation, but I'm having trouble with the part about $p$ inequivalent one-dim representations. The number of one dimensional representations is given by the order of the abelianization of the group: $$G/[G,G]$$ If the group is order $p$ or $p^2$ I can show that this holds, given that groups of order $p$ are cyclic, and therefore abelian, so the commutator subgroup is trivial. I also know that groups of order $p^2$ are abelian, so the commutator subgroup is again trivial.

But besides these cases I'm struggling to see how having a nontrivial center will yield the result. Any help would be appreciated!

Answer 1

Mark already indicated that $|G:G'|\geq p$ and that is the proof, but if you want to construct linear representations, you might throw in a bit more sophistication for example if you have a character table at your disposal. Let $H \lt G$ with $|G:H|=p$ (such a maximal subgroup exists). Look at $1_H^G$, the principal character of $H$ induced to $G$. By Frobenius' Reciprocity $[1_H^G,1_G]_G=[1_H,1_H]_H=1$, so $1_H^G=1_G + \sum_{\chi \in Irr(G)-\{1_G\}} a_{\chi}\chi$, with $a_{\chi}$ non-negative integers. Comparing degrees: $1_H^G(1)=|G:H|=p=1 + \sum_{\chi \in Irr(G)-\{1_G\}} a_{\chi}\chi(1)$. And we conclude that if $a_{\chi} \neq 0$ then $\chi(1) \lt p-1$. Since $\chi(1) \mid |G|$, we must have $\chi(1)=1$ for all $\chi$ and hence $1_H^G=1_G + \chi_2 + \cdots \chi_p$, with all $\chi_i$'s linear. And so here are your at least $p$ linear representations!