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.

  • $\begingroup$ There are no counterexamples of order up to and including 128. 256 seems like it would take about a day, and 512 is probably just silly. If you had more description of what the counterexample would look like, it might be possible to greatly speed up the search. $\endgroup$ – Jack Schmidt Jan 30 '11 at 5:15
  • $\begingroup$ I can now confirm that there are no counterexamples of order 256. But it would nice to have some suggestions as to why we might expect such a result to be true. $\endgroup$ – Derek Holt Jan 30 '11 at 11:05
  • $\begingroup$ The whole thing came from computations of capitulation kernels in subfields of the Hilbert class fields of quadratic number fields almost 20 years ago. An almost classical observation in this area is the fact that for quaternion extensions, the Schur multiplier is trivial and in each quadratic extension exactly one nontrivial class capitulates. Another conjecture from back then was shot down by Boston and Leedham-Green, so I figured I'd have just about the same luck in this case. I'd be deeply grateful if one of you could write up a description of the calculation (in sage/GAP?) as an answer $\endgroup$ – Franz Lemmermeyer Jan 30 '11 at 18:19

This is just a fairly literal translation of your question into GAP code.

If H is a subgroup of index 2 in G, and t is an element of G not in H, then the transfer map from G to H/[H,H] takes a particularly simple form:

If h in H, then h⋅1 = 1⋅h and h⋅t = t⋅ht, so tr(h) = h⋅ht.

If g in G∖H, then g⋅1 = t⋅(t−1g) and g⋅t = 1 ⋅(gt), so tr(g) = gt t−1g = g2.

In the GAP language, one creates the homomorphism with:

tr := function(g,h)
  local t, k, gen, img;
  t := First( g, t -> not t in h );
  k := MaximalAbelianQuotient(h);
  gen := GeneratorsOfGroup(g);
  img := List( gen, function(x)
    if x in h
    then return Image( k, x*x^t );
    else return Image( k, x*x );
  end );
  return GroupHomomorphismByImages( g, Image(k), gen, img );

We can test the conjecture on 2-groups with the following GAP function:

isCounter := g -> not ForAll( MaximalSubgroupClassReps( g ),
  h -> Index( Kernel( tr( g, h ) ), DerivedSubgroup( g ) )
  <= 2 * Product( AbelianInvariantsMultiplier( g ) ) );

We then look for examples of small orders with the following GAP function:

OneSmallGroup( [2,4,8,16,32], isCounter ); # 5 seconds

which simply returns fail since no such counterexample of order 2, 4, 8, 16, or 32 exists. To test the higher orders one uses:

OneSmallGroup( [ 64], isCounter ); # 30 seconds
OneSmallGroup( [128], isCounter ); # 10 minutes

For larger orders, one needs to be more careful calculating the Schur multiplier, or applying a little theory since the naive method is dealing with larger index subgroups and then larger integer matrices.


Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:

$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\rightarrow H_1(\Gamma,\tilde{\mathbb{Z}})$.

Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|\le 2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. Hatcher's Algebraic Topology textbook gives a long exact sequence for general coefficient systems (pg330), with $H_3G\stackrel{res}{\rightarrow}H_3\Gamma\rightarrow H_2(\Gamma,\tilde{\mathbb{Z}})\rightarrow H_2G\stackrel{res}{\rightarrow}H_2\Gamma$, so this could be of use.

  • Chris Gerig
  • $\begingroup$ But $\lvert \ker \operatorname{Ver}_{\Gamma\to G/G'}\rvert= 2$ and $M(\Gamma)=1$ for cyclic groups. So something must be wrong here, or am I missing something? $\endgroup$ – Frieder Ladisch Jan 31 '11 at 13:06
  • $\begingroup$ Ah there is an error: the $\Gamma$-action on the latter $\mathbb{Z}$ in the short exact sequence is not trivial, it acts as $\gamma\cdot z = -z$ for $\gamma$ not in $G$. I will make the fix, and hopefully this brings in the desired factor of 2. $\endgroup$ – Chris Gerig Jan 31 '11 at 18:41
  • $\begingroup$ Unfortunately the inequality $\lvert H_2(\Gamma,\tilde{\mathbb{Z}})\rvert \leq 2\cdot \lvert H_2(\Gamma)\rvert$ doesnt hold: Let $\Gamma = \text{SD}_{16} = \left<x,y | x^8, y^2, x^y=x^3 \right>$ and $G=\left< x^2, y\right> \cong D_8$ (an index $2$ subgroup). Then $H_1(\Gamma) \cong H_1(G) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2$ and the kernel $H_1(\Gamma) \xrightarrow{\text{tr}} H_1(G)$ has order $2$. The Schur multipliers equals $H_2(\Gamma)=0$ and $H_2(G)=\mathbb{Z}/2$. The long exact sequence then shows that $H_2(G,\tilde{\mathbb{Z}})$ has order $4$. $\endgroup$ – Kasper Andersen Jul 12 '18 at 9:54

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