3
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.