# Nilpotent groups are monomial

I'm trying to show that a nilpotent group $G$ is monomial; i.e., that every irreducible representation $\rho$ of $G$ satisfies $\rho = \text{Ind}_H^G(\tau)$ for some $H \leq G$, $\tau$ a one dimensional representation of $H$.

So far I've proved the claim by induction on $\left|G\right|$ in the case where $\ker\rho \neq \{1\}$. However, when $\rho$ is faithful, I'm not entirely sure where I should go...

I've also shown that if $G$ has a primitive faithful irreducible representation, then for any normal abelian subgroup $N \leq G$, we have $N \leq Z(G)$ and the question sheet gives the hint:

"any nilpotent group has a normal abelian self-centralising subgroup"

But I can't quite see how to tie these facts together and proceed. I know that we can pretty much reduce to the case of the representation being primitive by considering a minimal subgroup $H\leq G$ with the property that there exists $\tau \in \text{Irr}(H)$ with $\text{Ind}_H^G(\tau) = \rho$ - this $\tau$ is necessarily a primitive irreducible faithful representation of $H$. Otherwise I'm stumped. Any help would be much appreciated. Thanks!

• See Cor 23.4 for a clear proof of the title.
– zxx
Sep 10, 2022 at 18:56

Following the hints, you have an abelian normal subgroup $A\leq G$ that is self-centralising, i.e. $Z_G(A)=A$. But also, you are assuming that $G$ has a faithful primitive irreducible character, which forces $A$ to be contained in the centre of $G$, by the previous part of the exercise. What does it tell you that $A$ is in the centre, but is only centralised by itself?... Now you simply recall that $A$ was supposed to be abelian, so...
How about the proof given in Lemma $1$ by Ian D. Brown ? Another reference is given here.
• I don't 'know' anything about lower central series (second reference); I think I'm meant solely to solve it by considering the $\tau$ in the final paragraph of my question. I know that this $H$ is nilpotent and I'm then presumably meant to use some of the properties about nilpotent groups and their subgroups...but I can't seem to deduce anything useful May 11, 2014 at 13:41