For a $2$-group $P$ (that is, $|P|$ is a power of 2) let $K$ be the intersection of the kernels of the real-valued irreducible characters of $P.$ If the center $Z$ of $P$ is elementary abelian, then is $K$ abelian? GAP shows that it is when $|P|\le 256,$ though I suppose that is not particularly strong evidence.
To motivate the condition on the center, note that the Frattini subgroup $\Phi(Z)\subseteq K$ since each real-valued $\chi\in\mathrm{Irr}(G)$ lies over the identity or a sign character of $Z.$ Iterating, $E_i\subseteq K$ where $E_0=1$ and $E_i/E_{i-1}=\Phi(Z(P/E_{i-1}))$ for $i\ge 1.$ $E_\infty$ (and hence $K$) is not always abelian; my question is equivalent to asking if $K/E_\infty$ is abelian for any 2-group $P.$
As an perhaps slightly amusing aside, independent of my question above, by the Schur indicator formula, $P$ and $P/K$ (and hence also $P/E_\infty$) have the same numbers of involutions.
