Aspherical subcomplexes of finite aspherical 2-complexes

Question

Recently I have been considering the following question in low-dimensional topology, that would help with another result I have been working on, but have been assuming that the statement is likely too strong to be true. In particular it has certain similarities with the Whitehead conjecture https://en.m.wikipedia.org/wiki/Whitehead_conjecture, and hence I would predict it has a similar “unlikely to be true but hard to find a counterexample” nature.

Let a group $G$ be called aspherical if there is a finite aspherical 2-complex $X$ such $\pi_1(X)=G$.

Let $H\leqslant G$ be aspherical groups, does there exist some finite aspherical 2-complex $X$ with $\pi_1(X)=G$ that has an aspherical subcomplex with fundamental group $H$?

Although I expect the above question to be open and have not been able to make any headway on it, it is very possible I have missed a trick, so any solutions, pointers in the direction of a solution or reasons to believe it is untrue would be much appreciated.

I would also like to note that as this problem is parametrised by the group $G$ we can ask for solutions for specific aspherical groups. For example is this problem made easier if we look only at 1-relator groups?

Answer 1

Let $G=\mathbb{Z}\rtimes\mathbb{Z}$ be the fundamental group of the Klein bottle. It has a subgroup $H=\mathbb{Z}\rtimes 2\mathbb{Z}\cong \mathbb{Z}\times\mathbb{Z}$. If you like, view $H$ as the image of the fundamental group of the covering of the Klein bottle by the torus. Both $G$ and $H$ are clearly aspherical.

I claim that there does not exist a finite aspherical $2$-complex $X$ with aspherical subcomplex $Y\subseteq X$ such that $\pi_1(X)=G$ and $\pi_1(Y)=H$. Such an $X$ is homotopy equivalent to the Klein bottle, and such a $Y$ to the torus, so in integral cohomology with untwisted coefficients we have $$ H^2(X)\cong \mathbb{Z}/2,\qquad H^2(Y)\cong \mathbb{Z}. $$ By the long exact sequence of the pair $$ \cdots \to H^2(X)\to H^2(Y)\to H^3(X,Y)\to\cdots $$ we see that $H^3(X,Y)$ is nonzero, contradicting that the dimension of $X$ is at most $2$.