This is a long question, but hopefully someone can give me a suggestion, as I've been hitting my head against the wall...

Take a non-abelian group $P$, of order $p^3$, $p$ prime. We've already proven the following:

-$P$ has $p^2$ characters of degree 1, and $p-1$ irreducible characters of degree $p$. These are all the irreducible characters.

-$P$ has $p^2+p-1$ conjugacy classes. $p$ of these have size 1, $p^2-1$ have size $p$.

I still need to prove that: The classes of size $p$ are the non-identity cosets of the centre of $P$.

And that: given $\chi$, an irreducible character of degree $p$ $\implies$ The representation affording $\chi$ is faithful.

  • 2
    $\begingroup$ Since $G/Z(G)$ is abelian, the classes are all subsets of a single coset of $Z(G)$, so if they have the same size as those cosets, then they must be equal. The representation is faithful because if it had kernel $K$ then $G/K$ would be abelian, but all irreducible representations of abelian groups have degree $1$. $\endgroup$ – Derek Holt Nov 17 '14 at 5:02

Let $\chi \in \text {Irr}(G)$ with $\chi(1)=p$. Assume that the normal subgroup $\ker(\chi) \neq 1$. Since $G$ is a $p$-group it follows that $\text {ker}(\chi) \cap Z(G) \neq 1$. But $|Z(G)|=p$, hence we must have $Z(G) \subseteq \ker(\chi)$. Since $G$ is non-abelian of order $p^3$, we have $G'=Z(G)$ and it follows that $G' \subseteq \ker(\chi)$, which can only happen if $\chi$ is linear, a contradiction. So $\ker(\chi)=1$, that is $\chi$ is faithful.


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