Transfer kernels and the Schur multiplier

In this article we will prove that the capitulation kernel in cyclic unramified extensions is bounded in terms of the Schur multiplier of Galois groups. We include the relevant background information, and the actual proof takes up little more than a page. We begin with the history of the transfer map, present an elementary introduction to the Schur multiplier, describe the connection with the capitulation kernel in unramified abelian extensions of number fields, and give a proof of the conjecture in two special cases (of which the proof for groups with commutator groups $G'$ of order 2 is longer than the general proof), and finally give the proof of the main result, which actually takes up less than a page.

The Transfer Map

One of Hilbert’s conjectures on what we now call Hilbert class fields concerned the capitulation of ideal classes. If $k$ is a number field with Hilbert class field $k^1$, then the principal ideal theorem states that every ideal in $k$ becomes principal in $k^1$. More generally, let $K/k$ be an unramified abelian extension, let $K^1$ be the Hilbert class field of $K$, let $G=\operatorname{Gal}(K^1/k)$ be its Galois group over $k$, and let $H=\operatorname{Gal}(K^1/K)$. Since $K\subseteq k^1\subseteq K^1$, we have $G/G'\simeq\operatorname{Gal}(k^1/k)$. The map $\iota_{k\to K}:\operatorname{Cl}(k)\longrightarrow\operatorname{Cl}(K)$ induced by sending an ideal $\mathfrak a$ in $\mathcal O_k$ to $\mathfrak a\mathcal O_K$ is called the transfer of ideal classes; its kernel $\kappa_{K/k}=\ker\iota_{k\to K}$ consists of classes of ideals that become principal in $K$. Artin’s reciprocity law induces a corresponding homomorphism $\operatorname{Ver}$ (also called transfer) from $G/G'$ to $H$:

This allowed Artin to reduce the principal ideal theorem to a statement in group theory: If $G$ is a finite group, then the transfer map $\operatorname{Ver}:G/G'\longrightarrow G'/G''$ is the trivial map. The name transfer (Verlagerung) was introduced by Helmut Hasse in his report [6]. The transfer map is defined as follows. Let $H$ be a subgroup of finite index $n$ in a finite group $G$; write $G=Hg_1\cup\cdots\cup Hg_n$. For each $g\in G$ and each $i$ there is an $h_i\in H$ such that $g_i g=h_i g_{\sigma(i)}$ for a permutation $\sigma$ of $\{1,\ldots,n\}$. Then $t(g)=h_1h_2\cdots h_nH'$ defines a homomorphism $G\longrightarrow H/H'$ whose kernel contains $G'$. Thus $t$ induces a homomorphism

called the transfer map.

The transfer map was first used by Issai Schur in his article [13] published in 1902. It was Schur’s student Arnold Scholz who pointed out this fact in his Jahrbuch review (JFM 60.0895.01) of Iyanaga’s article [7] on the principal ideal theorem, which appeared in 1934 (contrary to the claims in [3 p. 166–167], it was not Iyanaga who had credited Schur with inventing the transfer map). Here’s what happened: Issai Schur wanted to publish the results of his PhD student Roberto Frucht in his journal, the Mathematische Zeitschrift. His co-editor Leon Lichtenstein objected to publishing a thesis in the journal and mentioned that Schur’s recent related results had appeared in Crelle’s journal. Thus Schur asked Hasse whether he would publish Frucht’s manuscript, and Hasse agreed to do that. When reading the submission (see [4]), Hasse noted similarities between some of Frucht’s formulas and those appearing in Furtwängler’s proof of the principal ideal theorem. He asked Schur whether he could explain these similarities, but Schur did not answer this question immediately.

When Artin visited Schur in September 1930 (Artin’s parents-in-law were living in Berlin), Artin worked on the proper definition of his $L$-series at ramified primes, and he talked with Schur about these things. Schur picked up Hasse’s report [6] and discovered that Artin’s transfer map in Furtwängler’s proof was the same map he had used in 1902. In a letter (see [15]) to Hasse written on December 30, 1930, he explained the connection in detail.

We do not know whether Scholz learned about Schur’s priority from Hasse or from his former teacher himself. Scholz used the transfer map after giving a talk in Hamburg in 1928, where he must have talked to Artin. Scholz later gave his student G. Hannink a topic on the transfer for his dissertation (Verlagerung und Nichteinfachheit von Gruppen, Kiel 1935).

Only in the 1970s was it discovered that the first example of the transfer map is Gauss’s Lemma in the third proof of the quadratic reciprocity law (see [12] and [17]).

The Schur multiplier

The results in this section are well known; see e.g. [9]. The Schur multiplier of a finite group $G$ is an important invariant that plays a major role in the investigation of class field towers. A group extension

is called a central extension if $A \subseteq Z(\Gamma)$ is contained in the center of $\Gamma$; in particular, $A$ is abelian. The group extension is called a stem extension if $A \subseteq\Gamma'$ is contained in the commutator group $\Gamma'$ of $\Gamma$. Schur has shown that for central stem extensions

the order of $A$ is bounded, and that the groups $A$ for which the order of $A$ is maximal are all isomorphic. In this case, $A=M(G)$ is called the Schur multiplier of $G$, and the group extension $D$ is called a Schur cover of $G$. The Klein 4-group $G=C_2 \times C_2$ has Schur multiplier $M(G) \cong C_2$, and its Schur cover groups are the dihedral group $D_8$ and the quaternionion group $Q_8$ of order 8.

In our proof below, we will use the following standard facts on Schur multipliers (see [9]).

(F0) $M(G) \cong H_2(G,\mathbb{Z})$. This is a finite abelian group annihilated by $|G|$.

(F1) (Hopf’s formula) If $G \cong F/R$ with $F$ free, then $M(G) \cong(R \cap F')/[F,R]$.

(F2) If $G \cong\bigoplus_{i=1}^{r} C_{m_i}$ is abelian, then

In particular, $M(G)=1$ if $G$ is cyclic.

(F3) If $1 \to A \to\Gamma\to G \to1$ is a central stem extension, then any set of lifts to $\Gamma$ of a generating set of $G$ generates $\Gamma$. In fact, if $Y$ is the subgroup generated by the lifts, then $\Gamma=YA$; as $A$ is central, $\Gamma'=Y'\subseteq Y$, hence $A\subseteq\Gamma'\subseteq Y$ and $\Gamma=Y$.

Next we prove a special case of the Ganea sequence (see [9]):

Proof. Write $G=F/R$ and let $S\supseteq R$ be the preimage of $Z$, so $G/Z=F/S$ and $S/R=Z$. By (F1), $M(G/Z)=(S\cap F')/[S,F]$, and $M(G)\to M(G/Z)$ is induced by the identity of $F$ (note $[R,F]\subseteq[S,F]$). Its image is $(R\cap F')[S,F]/[S,F]$, hence

Consider $\varphi:S\cap F'\to S/R[S,F]$. An element $rc$ with $r\in R$, $c\in[S,F]\subseteq F'$ lies in $F'$ if and only if $r\in F'$; so $\ker\varphi=(S\cap F')\cap R[S,F]=(R\cap F')[S,F]$, and $\operatorname{coker}\cong\operatorname{im}\varphi$. Since $Z$ is central, $[S,F]\subseteq R$, hence $R[S,F]=R$ and $S/R[S,F]=S/R=Z$. Finally $\operatorname{im}\varphi=\{fR:f\in F'\cap S\}=Z\cap F'R/R=Z\cap G'$ (if $f\in F'$ and $fR\in S/R$ then $f\in S$). □

Theorems of Furtwängler and Kisilevsky

Furtwängler [5] observed that if a number field $k$ has 2-class group of type $(2,2)$, then the 2-class group of its Hilbert 2-class field $k^1$ is trivial or cyclic; the 2-class field tower of $k$ then terminates at $k^1$ in the first case and at the second Hilbert 2-class field $k^2$ in the second. In addition, the Galois group of $k^2/k$ is either the Klein 4-group $(2,2)$, a dihedral group, a semi-dihedral group or a quaternion group. Kisilevsky [10] studied this situation in detail and showed that the Galois group $\operatorname{Gal}(k^2/k)$ is determined by the capitulation in the three unramified quadratic subextensions $k_1$, $k_2$ and $k_3$ of $k^1/k$. In the following, let $\kappa_j$ denote the capitulation kernel in $k_j/k$ and $C_j=N_{k_j/k}\operatorname{Cl}_2(k_j)\subset\operatorname{Cl}_2(k)$ the subgroup of ideal classes corresponding to $k_j/k$ via Artin’s reciprocity law. The following table contains the results of Furtwängler and Kisilevsky; in the last column we have added the order of the Schur multiplier of $G$:

$G = \operatorname{Gal}(k^{2}/k)$	capitulation	$|M(G)|$
$(2,2)$	$\kappa_{1} = \kappa_{2} = \kappa_{3} = \operatorname{Cl}_{2}(k)$	2
$Q_{8}$	$\kappa_{1} = C_{1}, \kappa_{2} = C_{2}, \kappa_{3} = C_{3}$	1
$D_{m}$	$\kappa_{1} = \operatorname{Cl}_{2}(k), \kappa_{2} = C_{3}, \kappa_{3} = C_{2}$	2
$\operatorname{SD}_{m}$	$\kappa_{1} \cap C_{1} = 1, \kappa_{2} = C_{3}, \kappa_{3} = C_{2}$	1
$Q_{m}, m \ge2^{4}$	$\kappa_{1} = C_{1}, \kappa_{2} = C_{3}, \kappa_{3} = C_{2}$	1

Here $D_{m}$, $Q_{m}$ and $\operatorname{SD}_{m}$ denote the dihedral, the quaternion and the semi-dihedral group of order $m = 2^{n}$. In the case $G = \operatorname{SD}_{m}$ we have $\kappa_{1} = C_{2}$, where $k_{2}$ is the quadratic subextension of $k^{1}$ for which $\operatorname{Gal}(k^{2}/k_{1})$ is a dihedral group.

One cannot fail to notice that if the Schur multiplier $M(G)$ is trivial, then $|\kappa_{j}| = 2$ for each quadratic unramified subextension $k_{j}$ of $k^{1}/k$. Computing capitulation kernels for other groups of order 16 with trivial Schur multiplier suggests the following conjecture:

We can formulate this conjecture without any reference to number fields by replacing the capitulation kernel with the kernel of the transfer map: If $G$ is a finite 2-group with trivial Schur multiplier, then $|\ker\operatorname{Ver}_{G/G'}\to H/H'| = 2$ for each subgroup $H$ of index 2 in $G$.

Computing the kernels of the transfer maps for 2-groups of small order with nontrivial Schur multiplier suggests that much more is true:

This conjecture goes back to [11 (7.20)]; there I have proved that if $k$ is a number field whose $p$-class field tower terminates with $k^{\infty}$ and where $G = \operatorname{Gal}(k^{\infty}/k)$, then for every cyclic unramified $p$-extension $K/k$ we have

Jack Schmidt and Derek Holt have verified this inequality for all 2-groups of order $\le256$.

The natural generalization of our conjecture to arbitrary $p$-groups is the following result, which is a special case of Theorem 7.

The more general conjecture that

for non-cyclic subgroups is not true: the groups [?, ?] and [?, ?] have Schur multiplier of order 2 and possess two subgroups of index 4 in which the kernel of the transfer map has order 16.

Simple Cases

Theorem 2 is easy to prove when the group $G$ is abelian:

A similar proof for groups with $|G'|=2$ requires considerably more work; we include it here because it involves classifying all 2-groups $G$ with $|G'|=2$ and trivial Schur multiplier.

Call a finite 2-group $G$ a $P$-group if $|G'|=2$ and $M(G)=1$. Let

Step 1: $\overline{G}/\overline{G}' \cong C_{2^a} \times C_2$. By Lemma 5(a), $|M(\overline{G})|=2$ for $\overline{G}=G/G'=\bigoplus_i C_{2^{a_i}}$ ($a_1 \ge\cdots\ge a_r \ge1$). By (F2), $\sum_{i<j}\min(a_i,a_j)=1$. This forces $r=2$, $a_2=1$ (for $r \le1$ the sum is 0; for $r \ge3$ it is $\ge3$). Put $a=a_1$; then $|G|=2^{a+2}$.

Step 2: generators and relations. Choose $x,y \in G$ whose images generate $\overline{G}$ with orders $2^a$ and 2. By (F3) (applied to the central stem extension $1 \to G' \to G \to\overline{G} \to1$), $G=\langle x,y\rangle$. The subgroup $G'$ is the normal closure of $z:=[x,y]$; being central of order 2, $G'=\langle z\rangle$, $z\ne1$, $z$ central. Also $x^{2^a},y^2\in G'$:

Note that $x$ has order $2^a$ and $\langle x\rangle\cap G'=1$ if $\alpha=0$ ($x^k\in G'$ forces $2^a\mid k$).

Step 3: $\alpha=0$ is impossible. We prove by induction on $a\ge1$: no $P$-group as in Step 2 has $\alpha=0$.

• $a=1$: then $x^2=1$ and $x\notin G'$, $|G|=8$, $|G'|=2$. $Q_8$ has a unique involution, which lies in $G'$; so $G\cong D_8$. But $|M(D_8)|=2$, so $D_8$ is no $P$-group.

• $a\ge2$: $x^2$ is central ($[x^2,y]=[x,y]^2=[x,y^2]=z^2=1$), hence $Z=\langle x^{2^{a-1}}\rangle\le\langle x^2\rangle$ is central of order 2, and $Z\cap G'\subseteq\langle x\rangle\cap G'=1$. By Lemma 5(b), $G/Z$ is a $P$-group. In $G/Z$ the images of $x,y$ generate, $\overline{x}^{2^{a-1}}=1$, $(G/Z)'=\langle z\rangle$, and the abelianization of $G/Z$ is $G/G'Z\cong C_{2^{a-1}}\times C_2$ with $x,y$ of orders $2^{a-1},2$. So $G/Z$ is a $P$-group of the type of Step 2 with parameter $a-1$ and $\alpha=0$, contradicting the induction hypothesis.

Step 4: $\alpha=1$. Now $z=x^{2^a}$, $x$ has order $2^{a+1}$, and $y^{-1}xy=xz=x^e$ with $e=1+2^a$. Put $n=a+2$; then $e=1+2^{n-2}$, $|G|=2^n$.

• $a=1$ ($n=3$): $x^4=1$, $y^{-1}xy=x^3=x^{-1}$. If $\beta=0$ this is $D_8$, excluded. If $\beta=1$ then $y^2=x^2$ and $G=Q_8$.

• $a\ge2$ ($n\ge4$): if $\beta=0$ then $x,y$ already satisfy the defining relations of $M_{2^n}$. If $\beta=1$ ($y^2=x^{2^a}$), put $y'=yx^{2^{a-1}}$. Since $x^k y=yx^{ek}$,

as $x^{2^{a+1}}=1$ and $2a-1\ge a+1$ for $a\ge2$; and $y'^{-1}xy'=x^e$. So again $x,y'$ satisfy the relations of $M_{2^n}$.

In the last case $G$ is a quotient of $M_{2^n}$ (generated by $x,y'$ satisfying its relations); since $|M_{2^n}|=2^n=|G|$ (a split extension of $\langle x\rangle\cong C_{2^{n-1}}$ by $C_2$, as $e^2\equiv1\pmod{2^{n-1}}$), $G\cong M_{2^n}$.

Proof of the Main Theorem

Let $G$ be a finite group and $H\lhd G$ a normal subgroup with $C=G/H=\langle\tau\rangle$ cyclic of order $p\ge2$. For a $p$-group $G$ and a subgroup $H$ of index $p$, this is automatic ($H$ is normal, and a group of prime order is cyclic).

Fix $t\in G$ and set $\tau=tH$. Since $G/H$ is abelian we have $G'\subseteq H$, and since $G/H=\langle tH\rangle$ we have $G=\langle H,t\rangle=\bigcup_{j=0}^{p-1}t^jH$. Put $A=H/H'$ (written additively unless stated otherwise). Since $A$ is abelian, $H$ acts trivially on $A$ by conjugation, so the conjugation action of $G$ on $A$ factors through $C$; this makes $A$ a $\mathbb{Z}[C]$-module, with $\overline{t}h=t^{-1}ht$ (independent of the choice of $t$). We write $A^\tau=\ker(\tau-1)$ for the group of $\tau$-fixed points and $N=1+\tau+\cdots+\tau^{p-1}\in\mathbb{Z}[C]$ for the norm element. Let $\operatorname{Ver}:G/G' \to A$ be the transfer and $M(G)=H_2(G,\mathbb{Z})$ the Schur multiplier of $G$; put $\kappa=\ker\operatorname{Ver}$ and $Q=\operatorname{coker}(M(H)\to M(G))$.

Proof. All three statements follow from $\lvert\ker\operatorname{Ver}\rvert=p\lvert Q\rvert$, since $Q$ is a quotient of $M(G)$ and $\lvert Q\rvert=1$ iff $M(H)\to M(G)$ is onto. □

Standard facts used

We use the following standard results on the homology groups $H_n(G,B)$

(H1) A short exact sequence of $G$-modules gives a long exact homology sequence;

(H2) Shapiro’s lemma: let $R=\mathbb{Z}[G/H]$ with $G$ acting by left translation. Regard $\mathbb{Z}$ as the trivial $H$-module; then $\operatorname{Ind}_H^G\mathbb{Z}=\mathbb{Z}[G]\otimes_{\mathbb{Z}[H]}\mathbb{Z}\cong R$ (induced and coinduced modules agree because $(G:H)<\infty$), and Shapiro’s lemma [1 III.(6.2)] gives natural isomorphisms $H_n(G,R)\cong H_n(H,\mathbb{Z})$ for all $n$. Under these isomorphisms,

• the augmentation $\varepsilon:\mathbb{Z}[G/H]\to\mathbb{Z}$ induces the map $i_*:H_n(H,\mathbb{Z})\to H_n(G,\mathbb{Z})$ induced by the inclusion $H\subseteq G$; [1 III.6]

• the map $\nu:\mathbb{Z}\to\mathbb{Z}[G/H]$, $1\mapsto\sum_{gH\in G/H}gH$, induces the transfer $H_n(G,\mathbb{Z})\to H_n(H,\mathbb{Z})$; [1 III.9]

• the $G$-endomorphism of $\mathbb{Z}[C]=\mathbb{Z}[G/H]$ given by multiplication by $\tau$ induces on $H_n(H,\mathbb{Z})$ the map $c\tau_*$ induced by conjugation by $t$; [1 III.8].

(H3) $H_0(G,\mathbb{Z})=\mathbb{Z}$; $H_1(H,\mathbb{Z})=H/H'$; $H_2(G,\mathbb{Z})=M(G)$ the Schur multiplier; on $H_1$ the transfer is the group-theoretic transfer $\operatorname{Ver}$; on $H_0$ the transfer $\mathbb{Z}\to\mathbb{Z}$ is multiplication by the index $(G:H)$ [1 III.9].

Two exact sequences

Let $R = \mathbb{Z}[C]$ with $C = G/H$, regarded as a $G$-module via $G \to C$, and let $\varepsilon: R \to\mathbb{Z}$, $\sum a_j\tau^j \mapsto\sum a_j$ be the augmentation with kernel $I$. Then $I = (\tau- 1)R$: the inclusion “$\supseteq$” is clear, and if $\sum a_j = 0$ then $\sum a_j\tau^j = \sum a_j(\tau^j - 1) = (\tau- 1)\sum_j a_j(1+\tau+\cdots+\tau^{j-1})$. Because $C$ is cyclic, the annihilator of $\tau- 1$ in $R$ is $\mathbb{Z}N$: indeed $(\tau- 1)\sum a_j\tau^j = \sum(a_{j-1}-a_j)\tau^j$ (indices mod $p$) vanishes iff all $a_j$ are equal. Hence $x \mapsto(\tau- 1)x$ gives a surjection $R \to I$ with kernel $\mathbb{Z}N$, and $\nu(1) = N$. We get

and $j\pi$ is multiplication by $\tau- 1$ on $R$. All maps are $G$-equivariant: $j$ is an inclusion of submodules; $\varepsilon(\tau\lambda) = \varepsilon(\lambda)$ (so $G$ acts trivially on $\mathbb{Z}$ in (I)); $\tau N = N$ (so $\nu$ is $G$-linear); and $\pi(\tau\lambda) = \tau(\tau- 1)\lambda= \tau\pi(\lambda)$ because $\mathbb{Z}[C]$ is commutative.

Sequence (I)

The long exact sequence attached to (I) contains

By (H2) and (H3) this reads

Exactness at $M(G)$ and at $H_1(G,I)$ shows that $\partial$ induces an isomorphism

Sequence (II)

The relevant part of the long exact homology sequence of (II) is

where we used $H_1(G,\mathbb{Z}) = G/G'$, $H_1(G,R) = H_1(H,\mathbb{Z}) = H/H'$ and $\nu_* = \operatorname{Ver}$ on $H_1$. Under $H_0(G,\mathbb{Z}) = \mathbb{Z}$ and $H_0(G,R) \cong H_0(H,\mathbb{Z}) = \mathbb{Z}$ (the latter induced by $\varepsilon$), the map $\nu_*$ is multiplication by $p$, since $\nu(1) = N \mapsto p$. Thus

Exactness at the fourth term gives $\operatorname{im}\partial' = 0$ since multiplication by $p$ is injective on $\mathbb{Z}$, so $\partial' = 0$, $\pi_*$ is onto, and exactness at $H/H'$ gives

Combining the two sequences

Let us collect what we have so far. Since $j\pi: R \longrightarrow I \longrightarrow R$ is multiplication by $\tau- 1$ on $R$, the induced map on $H_1(G,R) \cong A = H/H'$ is the endomorphism $j_*\pi_*: a \mapsto a^{\tau-1}$. Setting $A[\tau] = \ker(\tau- 1)$, the exact sequences extracted from the long exact homology sequences fit into the following exact and commutative diagram:

The commutativity of the left square follows from $\pi_*\operatorname{Ver}=0$ and $j_*\pi_*=\tau-1$. Applying the snake lemma then gives us an exact sequence

(which is easily proved by a simple diagram chase in this case), and in particular we have

Proof of Theorem 7

Proof. $L\subseteq G'$: For $h\in H$ we have, writing $A$ multiplicatively, $(\tau-1)\bar{h}=t^{-1}ht h^{-1}H'=[t,h^{-1}]H'$. Since $A$ is generated by the classes $\bar{h}$ and $\tau-1$ is a homomorphism, $(\tau-1)A$ is generated by these elements; hence $L$ is generated by $H'$ and the commutators $[t,h^{-1}]$, $h\in H$. All of these lie in $G'$, so $L\subseteq G'$.

$L$ is normal in $G$: The subgroup $(\tau-1)A$ is a $\mathbb{Z}[C]$-submodule of $A$, because $\tau$ commutes with $\tau-1$. The conjugation action of $G$ on $A$ factors through $C$, hence $(\tau-1)A$ is stable under all of $G$. So for $g\in G$ and $x\in L$ we have $g^{-1}xg\in H$ (as $H\trianglelefteq G$) and its class in $A$ lies in $(\tau-1)A$, i.e. $g^{-1}xg\in L$.

$G'\subseteq L$: The image of $H$ in $G/L$ is abelian since $H'\subseteq L$. For $h\in H$ we have $[t,h^{-1}]\in L$, i.e. $t^{-1}ht=h$ in $G/L$, so $t$ commutes with the image of every element of $H$. Since $G=\langle H,t\rangle$, the group $G/L$ is generated by the images of $H$ and $t$, which commute pairwise; hence $G/L$ is abelian and $G'\subseteq L$.

Counting: We have shown that $G'=L$, so $(H:G')=|A/(\tau-1)A|=|A^\tau|$, because kernel and cokernel of an endomorphism of a finite group have the same order. Since $G'\subseteq H$ we get $|G:G'|=|G:H||H:G'|=p|A^\tau|$. □

Proof of Theorem 7. The exact sequence $0\to\kappa\to G/G'\to A^\tau\to Q\to0$ was established in (5). Hence $|G/G'|=|\kappa||\operatorname{Ver}(G/G')|$ and $|Q|=|A^\tau|/|\operatorname{Ver}(G/G')|$. By Proposition 10,

The cokernel is a quotient of $\mathrm{M}(G)$, giving the inequality, with equality iff the cokernel is all of $\mathrm{M}(G)$, i.e. iff $\mathrm{M}(H)\to\mathrm{M}(G)$ is zero. □

Remarks

In November 2025 I looked at Folk’s article [2]. He said that there is extra capitulation in a cyclic unramified extension $K/k$ if the order of $\kappa_{K/k}$ is strictly larger than $(K:k)$. He showed that extra capitulation implies the existence of certain central extensions, which in turn, so I hoped, would show that the Schur multiplier of the Galois group of $K^1/k$ cannot be trivial. Before I found the time to think this through, the AI revolution happened.

Now that the transfer conjecture is proved we might ask conversely how these field extensions corresponding to the cokernel of $M(H) \longrightarrow M(G)$ may be constructed.

Acknowledgement

The proof was found with the help of the free version of Claude in October 2026. Over several sessions, Claude suggested three proofs: the first used spectral sequences, the second replaced everything by explicit calculations based on Hopf’s formula; I chose to present the third version using standard homology groups in my own exposition.

I would like to dedicate this article to the memory of my dear friend Chip Snyder (1946–2026), who kept returning to these questions during the final years of his life. In particular, together with Elliot Benjamin (see [16]) he proved that the transfer conjecture holds in the following case: If $G$ is a finite $2$-group and $H$ is a maximal abelian subgroup of $G$, then $\operatorname{rk} \ker\operatorname{Ver}(G/G^\prime\longrightarrow H/H^\prime) \le1 + \operatorname{rk} M(G)$.