$g \in G$ acts trivially on all irreducible representations of $G/N$

Question

There is a one-to-one correspondence between irreducible representations of $G$ on which $N$ acts trivially and irreducible representations of $G/N$. Suppose we have some $g \in G$ which acts trivially on all irreducible representations on which $N$ acts trivially. We wish to show that $g$ must, in fact, belong to $N$.

My first thought was to find a particular representation which would force $g \in N$ giving the result but that is silly as that would imply a much broader result (and we don't really have access to what the representations look like). So, given that, my thought is to show that $g$ must, in fact, conjugate into $N$ (forcing $g \in N$) which is to say the conjugacy class of $g$ in $G$ nontrivially intersects $N$ since characters agree on conjugacy classes but I've hit a bit of a roadblock for the moment. I was hoping someone could provide some guidance. Thanks!

Answer 1

Looking at it from a contra-positive perspective you want to show that if $g \in G$ does not belong to $N$, there must be an irreducible representation $V$ on which $g$ acts non-trivially, while $N$ does act trivially.

This statement sounds a bit awkward, but luckily you already mention the correspondence between irreps of $G$ on which $N$ acts trivially and irreps of $G/N$. Let $h$ be the equivalence class of $g$ in $H := G/N$. The non-awkward sounding version of what we want to prove is then:

If $h \in H$ is not equal to the identity, then there exists an irreducible representation $V$ of $H$ such that $h$ acts non-trivially on $V$.

It is indeed not hard to see that if we can find such a $V$, then the representation of the full group $G$ on the same vector space given by the one-to-one correspondence does indeed do what we want. So that is nice. But there is something even nicer. Suppose someone came in right now and only read the statement in yellow, this person would have no idea that we think about $H$ as $G/N$ or even have any reason to ponder the existence of either $G$ or $N$ at all. Instead they could just think 'Ha, nice statement about a group...' (just one group!) '...let's see if it is true'. And in the end they could come to the conclusion that it is. (Ehm, at least for finite groups. Maybe among infinite groups this is not true, I am not sure there.)

Perhaps you want to stare at the yellow statement and try and prove it yourself, so I won't tell you how to do it, but if you want a further hint, just let me know and I'll cook one up.