Let $\Gamma$ be a finite $2$-group, and let $G$ be any subgroup of index $2$. Moreover, let Ver$: \Gamma/\Gamma' \to G/G'$ denote the group theoretical transfer, and let $M(\Gamma)$ be the Schur multiplier of $\Gamma$.

Is it true that $$ | \text{ker}\ {\rm Ver}_{\Gamma/\Gamma' \to G/G'} | \le 2 | M(\Gamma) | \quad ? $$

The answer is positive for a couple of groups with small multiplier, such as cyclic $2$-groups, which have trivial Schur multiplier, the groups of order $8$, or dihedral and quaternion groups. Does current computer technology allow finding a counterexample by going through 2-groups of moderate size?

**Edit.** Here's a little bit of background. Let $K$ be a quadratic number field whose 2-class group has type (2,2). It is known that the Hilbert 2-class field $K^1$ has cyclic
2-class group, and that the Galois group $\Gamma$ of the second Hilbert 2-class field is either (2,2) itself, a dihedral, quaternion, or semi-dihedral 2-group. Let $K_j/K$ (j=1, 2, 3) denote the three unramified quadratic extensions inside $K^1$, and let $G_j$ be the
Galois groups of $K^2/K_j$. Then $\Gamma$ is quaternion or semi-dihedral if and only if
in each of the extensions $K_j/K$, exactly one ideal class of order 2 capitulates (i.e., becomes principal). Now these are exactly the groups among the possible Galois groups with trivial Schur multiplier. On the other hand, the transfer of ideal classes corresponds, via Artin's reciprocity law, to the transfer map (Verlagerung) from the abelianization of $\Gamma$ to that of the $G_j$. Thus in this case, we find that the order of the Schur multiplier is equal to one half of the maximal order of the capitulation kernels.