Finite group is Dedekind iff for every irrep, every element acts as identity or has all eigenvalues $\ne 1$

Question

Consider the following claim: a finite group $G$ is Dedekind $\iff$ for every irrep $\rho$, and every $g \in G$, $\rho(g)$ either is identity matrix or has all eigenvalues $\ne 1$.

Is this claim true?

Recall that finite Dedekind groups are abelian or $Q_8 \times \mathbb{Z}_2^t \times \mathbb{Z}_m$ for odd $m$, integer $t$, $Q_8$ the quaternion group.

Dedekind $\implies$ for every irrep $\rho$, and every $g \in G$, $\rho(g)$ either is identity matrix or has all eigenvalues $\ne 1$ holds by inspection. So the question is whether $\impliedby$ holds.

Answer 1

The condition can be restated as: for every $g$ and $\rho$, the subspace $\mathrm{Ker}(\rho(g)-1)$ is a subrepresentation.

If $G$ is Dedekind (meaning that every subgroup is normal, or equivalently that every cyclic subgroup is normal), then $\langle g\rangle$ being normal, the result follows (no need to use inspection).

Conversely, suppose that $\mathrm{Ker}(\lambda(g)-1)$ is a subrepresentation for $\lambda$ the left regular representation. This kernel is the set $U_g$ of functions that are left $\langle g\rangle$-invariant, and we readily see that for $h\in G$, $hU_g=U_{hgh^{-1}}$, and $U_g=U_{g'}$ iff $\langle g\rangle =\langle g'\rangle$. So the condition that $U_g$ is a subrepresentation means that $\langle g\rangle$ is normal. If this is true for every $g$, this means that $G$ is Dedekind.