Artin transfer (group theory)

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In the mathematical field of group theory, an Artin transfer is a certain homomorphism from an arbitrary finite or infinite group to the commutator quotient group of a subgroup of finite index. Originally, such mappings arose as group theoretic counterparts of class extension homomorphisms of abelian extensions of algebraic number fields by applying Artin's reciprocity maps to ideal class groups and analyzing the resulting homomorphisms between quotients of Galois groups. However, independently of number theoretic applications, a partial order on the kernels and targets of Artin transfers has recently turned out to be compatible with parent-descendant relations between finite p-groups (with a prime number p), which can be visualized in descendant trees. Therefore, Artin transfers provide a valuable tool for the classification of finite p-groups and for searching and identifying particular groups in descendant trees by looking for patterns defined by the kernels and targets of Artin transfers. These strategies of pattern recognition are useful in purely group theoretic context, as well as for applications in algebraic number theory concerning Galois groups of higher p-class fields and Hilbert p-class field towers.

Let ${\displaystyle G}$ be a group and ${\displaystyle H\leq G}$ be a subgroup of finite index ${\displaystyle n.}$

Definitions.^[1] A left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ is an ordered system ${\displaystyle (g_{1},\ldots ,g_{n})}$ of representatives for the left cosets of ${\displaystyle H}$ in ${\displaystyle G}$ such that

${\displaystyle G=\bigsqcup _{i=1}^{n}g_{i}H.}$

Similarly a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$ is an ordered system ${\displaystyle (d_{1},\ldots ,d_{n})}$ of representatives for the right cosets of ${\displaystyle H}$ in ${\displaystyle G}$ such that

${\displaystyle G=\bigsqcup _{i=1}^{n}Hd_{i}.}$

Remark. For any transversal of ${\displaystyle H}$ in ${\displaystyle G}$, there exists a unique subscript ${\displaystyle 1\leq i_{0}\leq n}$ such that ${\displaystyle g_{i_{0}}\in H}$, resp. ${\displaystyle d_{i_{0}}\in H}$. Of course, this element with subscript ${\displaystyle i_{0}}$ which represents the principal coset (i.e., the subgroup ${\displaystyle H}$ itself) may be, but need not be, replaced by the neutral element ${\displaystyle 1}$.

Lemma.^[2] Let ${\displaystyle G}$ be a non-abelian group with subgroup ${\displaystyle H}$. Then the inverse elements ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ of a left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$ of ${\displaystyle H}$ in ${\displaystyle G}$ form a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$. Moreover, if ${\displaystyle H}$ is a normal subgroup of ${\displaystyle G}$, then any left transversal is also a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$.

Proof. Since the mapping ${\displaystyle x\mapsto x^{-1}}$ is an involution of ${\displaystyle G}$ we see that:

${\displaystyle G=G^{-1}=\bigsqcup _{i=1}^{n}(g_{i}H)^{-1}=\bigsqcup _{i=1}^{n}H^{-1}g_{i}^{-1}=\bigsqcup _{i=1}^{n}Hg_{i}^{-1}.}$

For a normal subgroup ${\displaystyle H}$ we have ${\displaystyle xH=Hx}$ for each ${\displaystyle x\in G}$.

We must check when the image of a transversal under a homomorphism is also a transversal.

Proposition. Let ${\displaystyle \phi :G\to K}$ be a group homomorphism and ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of a subgroup ${\displaystyle H}$ in ${\displaystyle G}$ with finite index ${\displaystyle n.}$ The following two conditions are equivalent:

• ${\displaystyle (\phi (g_{1}),\ldots ,\phi (g_{n}))}$ is a left transversal of the subgroup ${\displaystyle \phi (H)}$ in the image ${\displaystyle \phi (G)}$ with finite index ${\displaystyle (\phi (G):\phi (H))=n.}$
• ${\displaystyle \ker(\phi )\leq H.}$

Proof. As a mapping of sets ${\displaystyle \phi }$ maps the union to another union:

${\displaystyle \phi (G)=\phi \left(\bigcup _{i=1}^{n}g_{i}H\right)=\bigcup _{i=1}^{n}\phi (g_{i}H)=\bigcup _{i=1}^{n}\phi (g_{i})\phi (H),}$

but weakens the equality for the intersection to a trivial inclusion:

${\displaystyle \emptyset =\phi (\emptyset )=\phi (g_{i}H\cap g_{j}H)\subseteq \phi (g_{i}H)\cap \phi (g_{j}H)=\phi (g_{i})\phi (H)\cap \phi (g_{j})\phi (H),\qquad i\neq j.}$

Suppose for some ${\displaystyle 1\leq i\leq j\leq n}$:

${\displaystyle \phi (g_{i})\phi (H)\cap \phi (g_{j})\phi (H)\neq \emptyset }$

then there exists elements ${\displaystyle h_{i},h_{j}\in H}$ such that

${\displaystyle \phi (g_{i})\phi (h_{i})=\phi (g_{j})\phi (h_{j})}$

Then we have:

${\displaystyle {\begin{aligned}\phi (g_{i})\phi (h_{i})=\phi (g_{j})\phi (h_{j})&\Longrightarrow \phi (g_{j})^{-1}\phi (g_{i})\phi (h_{i})\phi (h_{j})^{-1}=1\\&\Longrightarrow \phi \left(g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\right)=1\\&\Longrightarrow g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\in \ker(\phi )\\&\Longrightarrow g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\in H&&\ker(\phi )\leq H\\&\Longrightarrow g_{j}^{-1}g_{i}\in H&&h_{i}h_{j}^{-1}\in H\\&\Longrightarrow g_{i}H=g_{j}H\\&\Longrightarrow i=j\end{aligned}}}$

Conversely if ${\displaystyle \ker(\phi )\nsubseteq H}$ then there exists ${\displaystyle x\in G\setminus H}$ such that ${\displaystyle \phi (x)=1.}$ But the homomorphism ${\displaystyle \phi }$ maps the disjoint cosets ${\displaystyle x\cdot H\cap 1\cdot H=\emptyset }$ to equal cosets:

${\displaystyle \phi (x)\phi (H)\cap \phi (1)\phi (H)=1\cdot \phi (H)\cap 1\cdot \phi (H)=\phi (H).}$

Remark. We emphasize the important equivalence of the proposition in a formula:

${\displaystyle (1)\quad \ker(\phi )\leq H\quad \Longleftrightarrow \quad {\begin{cases}\phi (G)=\bigsqcup _{i=1}^{n}\phi (g_{i})\phi (H)\\(\phi (G):\phi (H))=n\end{cases}}}$

Suppose ${\displaystyle (g_{1},\ldots ,g_{n})}$ is a left transversal of a subgroup ${\displaystyle H}$ of finite index ${\displaystyle n}$ in a group ${\displaystyle G}$. A fixed element ${\displaystyle x\in G}$ gives rise to a unique permutation ${\displaystyle \pi _{x}\in S_{n}}$ of the left cosets of ${\displaystyle H}$ in ${\displaystyle G}$ by left multiplication such that:

${\displaystyle (2)\quad \forall i\in \{1,\ldots ,n\}:\qquad xg_{i}H=g_{\pi _{x}(i)}H\Longrightarrow xg_{i}\in g_{\pi _{x}(i)}H.}$

Using this we define a set of elements called the monomials associated with ${\displaystyle x}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H.}$

Similarly, if ${\displaystyle (d_{1},\ldots ,d_{n})}$ is a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$, then a fixed element ${\displaystyle x\in G}$ gives rise to a unique permutation ${\displaystyle \rho _{x}\in S_{n}}$ of the right cosets of ${\displaystyle H}$ in ${\displaystyle G}$ by right multiplication such that:

${\displaystyle (3)\quad \forall i\in \{1,\ldots ,n\}:\qquad Hd_{i}x=Hd_{\rho _{x}(i)}\Longrightarrow d_{i}x\in Hd_{\rho _{x}(i)}.}$

And we define the monomials associated with ${\displaystyle x}$ with respect to ${\displaystyle (d_{1},\ldots ,d_{n})}$:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H.}$

Definition.^[1] The mappings:

${\displaystyle {\begin{cases}G\to S_{n}\\x\mapsto \pi _{x}\end{cases}}\qquad {\begin{cases}G\to S_{n}\\x\mapsto \rho _{x}\end{cases}}}$

are called the permutation representation of ${\displaystyle G}$ in the symmetric group ${\displaystyle S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ and ${\displaystyle (d_{1},\ldots ,d_{n})}$ respectively.

Definition.^[1] The mappings:

${\displaystyle {\begin{cases}G\to H^{n}\times S_{n}\\x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\end{cases}}\qquad {\begin{cases}G\to H^{n}\times S_{n}\\x\mapsto (w_{x}(1),\ldots ,w_{x}(n);\rho _{x})\end{cases}}}$

are called the monomial representation of ${\displaystyle G}$ in ${\displaystyle H^{n}\times S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ and ${\displaystyle (d_{1},\ldots ,d_{n})}$ respectively.

Lemma. For the right transversal ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ associated to the left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$, we have the following relations between the monomials and permutations corresponding to an element ${\displaystyle x\in G}$:

${\displaystyle (4)\quad {\begin{cases}w_{x^{-1}}(i)=u_{x}(i)^{-1}&1\leq i\leq n\\\rho _{x^{-1}}=\pi _{x}\end{cases}}}$

Proof. For the right transversal ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$, we have ${\displaystyle w_{x}(i)=g_{i}^{-1}xg_{\rho _{x}(i)}}$, for each ${\displaystyle 1\leq i\leq n}$. On the other hand, for the left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$, we have

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i)^{-1}=\left(g_{\pi _{x}(i)}^{-1}xg_{i}\right)^{-1}=g_{i}^{-1}x^{-1}g_{\pi _{x}(i)}=g_{i}^{-1}x^{-1}g_{\rho _{x^{-1}}(i)}=w_{x^{-1}}(i).}$

This relation simultaneously shows that, for any ${\displaystyle x\in G}$, the permutation representations and the associated monomials are connected by ${\displaystyle \rho _{x^{-1}}=\pi _{x}}$ and ${\displaystyle w_{x^{-1}}(i)=u_{x}(i)^{-1}}$ for each ${\displaystyle 1\leq i\leq n}$.

Definitions.^[2][3] Let ${\displaystyle G}$ be a group and ${\displaystyle H}$ a subgroup of finite index ${\displaystyle n.}$ Assume ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ is a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ with associated permutation representation ${\displaystyle \pi _{x}:G\to S_{n},}$ such that

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H.}$

Similarly let ${\displaystyle (d)=(d_{1},\ldots ,d_{n})}$ be a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$ with associated permutation representation ${\displaystyle \rho _{x}:G\to S_{n}}$ such that

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H.}$

The Artin transfer ${\displaystyle T_{G,H}^{(g)}:G\to H/H'}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ is defined as:

${\displaystyle (5)\quad \forall x\in G:\qquad T_{G,H}^{(g)}(x):=\prod _{i=1}^{n}g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot H'.}$

Similarly we define:

${\displaystyle (6)\quad \forall x\in G:\qquad T_{G,H}^{(d)}(x):=\prod _{i=1}^{n}d_{i}xd_{\rho _{x}(i)}^{-1}\cdot H'=\prod _{i=1}^{n}w_{x}(i)\cdot H'.}$

Remarks. Isaacs^[4] calls the mappings

${\displaystyle {\begin{cases}P:G\to H\\x\mapsto \prod _{i=1}^{n}u_{x}(i)\end{cases}}\qquad {\begin{cases}P:G\to H\\x\mapsto \prod _{i=1}^{n}w_{x}(i)\end{cases}}}$

the pre-transfer from ${\displaystyle G}$ to ${\displaystyle H}$. The pre-transfer can be composed with a homomorphism ${\displaystyle \phi :H\to A}$ from ${\displaystyle H}$ into an abelian group ${\displaystyle A}$ to define a more general version of the transfer from ${\displaystyle G}$ to ${\displaystyle A}$ via ${\displaystyle \phi }$, which occurs in the book by Gorenstein.^[5]

${\displaystyle {\begin{cases}(\phi \circ P):G\to A\\x\mapsto \prod _{i=1}^{n}\phi (u_{x}(i))\end{cases}}\qquad {\begin{cases}(\phi \circ P):G\to A\\x\mapsto \prod _{i=1}^{n}\phi (w_{x}(i))\end{cases}}}$

Taking the natural epimorphism

${\displaystyle {\begin{cases}\phi :H\to H/H'\\v\mapsto vH'\end{cases}}}$

yields the preceding definition of the Artin transfer ${\displaystyle T_{G,H}}$ in its original form by Schur^[2] and by Emil Artin,^[3] which has also been dubbed Verlagerung by Hasse.^[6] Note that, in general, the pre-transfer is neither independent of the transversal nor a group homomorphism.

Proposition.^{[1][2][4][5][7][8][9]} The Artin transfers with respect to any two left transversals of ${\displaystyle H}$ in ${\displaystyle G}$ coincide.

Proof. Let ${\displaystyle (\ell )=(\ell _{1},\ldots ,\ell _{n})}$ and ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ be two left transversals of ${\displaystyle H}$ in ${\displaystyle G}$. Then there exists a unique permutation ${\displaystyle \sigma \in S_{n}}$ such that:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad g_{i}H=\ell _{\sigma (i)}H.}$

Consequently:

${\displaystyle \forall i\in \{1,\ldots ,n\},\exists h_{i}\in H:\qquad g_{i}h_{i}=\ell _{\sigma (i)}.}$

For a fixed element ${\displaystyle x\in G}$, there exists a unique permutation ${\displaystyle \lambda _{x}\in S_{n}}$ such that:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad \ell _{\lambda _{x}(\sigma (i))}H=x\ell _{\sigma (i)}H=xg_{i}h_{i}H=xg_{i}H=g_{\pi _{x}(i)}H=g_{\pi _{x}(i)}h_{\pi _{x}(i)}H=\ell _{\sigma (\pi _{x}(i))}H.}$

Therefore, the permutation representation of ${\displaystyle G}$ with respect to ${\displaystyle (\ell _{1},\ldots ,\ell _{n})}$ is given by ${\displaystyle \lambda _{x}\circ \sigma =\sigma \circ \pi _{x}}$ which yields: ${\displaystyle \lambda _{x}=\sigma \circ \pi _{x}\circ \sigma ^{-1}\in S_{n}.}$ Furthermore, for the connection between the two elements:

${\displaystyle {\begin{aligned}v_{x}(i)&:=\ell _{\lambda _{x}(i)}^{-1}x\ell _{i}\in H\\u_{x}(i)&:=g_{\pi _{x}(i)}^{-1}xg_{i}\in H\end{aligned}}}$

we have:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad v_{x}(\sigma (i))=\ell _{\lambda _{x}(\sigma (i))}^{-1}x\ell _{\sigma (i)}=\ell _{\sigma (\pi _{x}(i))}^{-1}xg_{i}h_{i}=\left(g_{\pi _{x}(i)}h_{\pi _{x}(i)}\right)^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}.}$

Finally since ${\displaystyle H/H'}$ is abelian and ${\displaystyle \sigma }$ and ${\displaystyle \pi _{x}}$ are permutations, the Artin transfer turns out to be independent of the left transversal:

${\displaystyle T_{G,H}^{(\ell )}(x)=\prod _{i=1}^{n}v_{x}(\sigma (i))\cdot H'=\prod _{i=1}^{n}h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\prod _{i=1}^{n}h_{\pi _{x}(i)}^{-1}\prod _{i=1}^{n}h_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot 1\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot H'=T_{G,H}^{(g)}(x),}$

as defined in formula (5).

Proposition. The Artin transfers with respect to any two right transversals of ${\displaystyle H}$ in ${\displaystyle G}$ coincide.

Proof. Similar to the previous proposition.

Proposition. The Artin transfers with respect to ${\displaystyle (g^{-1})=(g_{1}^{-1},\ldots ,g_{n}^{-1})}$ and ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ coincide.

Proof. Using formula (4) and ${\displaystyle H/H'}$ being abelian we have:

${\displaystyle T_{G,H}^{(g^{-1})}(x)=\prod _{i=1}^{n}g_{i}^{-1}xg_{\rho _{x}(i)}\cdot H'=\prod _{i=1}^{n}w_{x}(i)\cdot H'=\prod _{i=1}^{n}u_{x^{-1}}(i)^{-1}\cdot H'=\left(\prod _{i=1}^{n}u_{x^{-1}}(i)\cdot H'\right)^{-1}=\left(T_{G,H}^{(g)}\left(x^{-1}\right)\right)^{-1}=T_{G,H}^{(g)}(x).}$

The last step is justified by the fact that the Artin transfer is a homomorphism. This will be shown in the following section.

Corollary. The Artin transfer is independent of the choice of transversals and only depends on ${\displaystyle H}$ and ${\displaystyle G}$.

Theorem.^{[1][2][4][5][7][8][9]} Let ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$. The Artin transfer

${\displaystyle {\begin{cases}T_{G,H}:G\to H/H'\\x\mapsto \prod _{i=1}^{n}g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H'\end{cases}}}$

and the permutation representation:

${\displaystyle {\begin{cases}G\to S_{n}\\x\mapsto \pi _{x}\end{cases}}}$

are group homomorphisms:

${\displaystyle (7)\quad \forall x,y\in G:\qquad T_{G,H}(xy)=T_{G,H}(x)\cdot T_{G,H}(y)\quad {\text{and}}\quad \pi _{xy}=\pi _{x}\circ \pi _{y}.}$

It is illuminating to restate the homomorphism property of the Artin transfer in terms of the monomial representation. The images of the factors ${\displaystyle x,y}$ are given by

${\displaystyle T_{G,H}(x)=\prod _{i=1}^{n}u_{x}(i)\cdot H'\quad {\text{and}}\quad T_{G,H}(y)=\prod _{j=1}^{n}u_{y}(j)\cdot H'.}$

In the last proof, the image of the product ${\displaystyle xy}$ turned out to be

${\displaystyle T_{G,H}(xy)=\prod _{j=1}^{n}g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H'=\prod _{j=1}^{n}u_{x}(\pi _{y}(j))\cdot u_{y}(j)\cdot H'}$,

which is a very peculiar law of composition discussed in more detail in the following section.

The law is reminiscent of crossed homomorphisms ${\displaystyle x\mapsto u_{x}}$ in the first cohomology group ${\displaystyle \mathrm {H} ^{1}(G,M)}$ of a ${\displaystyle G}$-module ${\displaystyle M}$, which have the property ${\displaystyle u_{xy}=u_{x}^{y}\cdot u_{y}}$ for ${\displaystyle x,y\in G}$.

The peculiar structures which arose in the previous section can also be interpreted by endowing the cartesian product ${\displaystyle H^{n}\times S_{n}}$ with a special law of composition known as the wreath product ${\displaystyle H\wr S_{n}}$ of the groups ${\displaystyle H}$ and ${\displaystyle S_{n}}$ with respect to the set ${\displaystyle \{1,\ldots ,n\}.}$

Definition. For ${\displaystyle x,y\in G}$, the wreath product of the associated monomials and permutations is given by

${\displaystyle (8)\quad (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\cdot (u_{y}(1),\ldots ,u_{y}(n);\pi _{y}):=(u_{x}(\pi _{y}(1))\cdot u_{y}(1),\ldots ,u_{x}(\pi _{y}(n))\cdot u_{y}(n);\pi _{x}\circ \pi _{y})=(u_{xy}(1),\ldots ,u_{xy}(n);\pi _{xy}).}$

Theorem.^[1][7] With this law of composition on ${\displaystyle H^{n}\times S_{n}}$ the monomial representation

${\displaystyle {\begin{cases}G\to H\wr S_{n}\\x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\end{cases}}}$

is an injective homomorphism.

Remark. The monomial representation of the theorem stands in contrast to the permutation representation, which cannot be injective if ${\displaystyle |G|>n!.}$

Remark. Whereas Huppert^[1] uses the monomial representation for defining the Artin transfer, we prefer to give the immediate definitions in formulas (5) and (6) and to merely illustrate the homomorphism property of the Artin transfer with the aid of the monomial representation.

Theorem.^[1][7] Let ${\displaystyle G}$ be a group with nested subgroups ${\displaystyle K\leq H\leq G}$ such that ${\displaystyle (G:H)=n,(H:K)=m}$ and ${\displaystyle (G:K)=(G:H)\cdot (H:K)=nm<\infty .}$ Then the Artin transfer ${\displaystyle T_{G,K}}$ is the compositum of the induced transfer ${\displaystyle {\tilde {T}}_{H,K}:H/H'\to K/K'}$ and the Artin transfer ${\displaystyle T_{G,H}}$, that is:

${\displaystyle (9)\quad T_{G,K}={\tilde {T}}_{H,K}\circ T_{G,H}}$.

Finally, we want to emphasize the structural peculiarity of the monomial representation

${\displaystyle {\begin{cases}G\to K^{n\cdot m}\times S_{n\cdot m}\\x\mapsto (k_{x}(1,1),\ldots ,k_{x}(n,m);\gamma _{x})\end{cases}}}$

which corresponds to the composite of Artin transfers, defining

${\displaystyle k_{x}(i,j):=\left((gh)_{\gamma _{x}(i,j)}\right)^{-1}x(gh)_{(i,j)}\in K}$

for a permutation ${\displaystyle \gamma _{x}\in S_{n\cdot m}}$, and using the symbolic notation ${\displaystyle (gh)_{(i,j)}:=g_{i}h_{j}}$ for all pairs of subscripts ${\displaystyle 1\leq i\leq n}$, ${\displaystyle 1\leq j\leq m}$.

The preceding proof has shown that

${\displaystyle k_{x}(i,j)=h_{\sigma _{y_{i}}(j)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{j}.}$

Therefore, the action of the permutation ${\displaystyle \gamma _{x}}$ on the set ${\displaystyle [1,n]\times [1,m]}$ is given by ${\displaystyle \gamma _{x}(i,j)=(\pi _{x}(i),\sigma _{u_{x}(i)}(j))}$. The action on the second component ${\displaystyle j}$ depends on the first component ${\displaystyle i}$ (via the permutation ${\displaystyle \sigma _{u_{x}(i)}\in S_{m}}$), whereas the action on the first component ${\displaystyle i}$ is independent of the second component ${\displaystyle j}$. Therefore, the permutation ${\displaystyle \gamma _{x}\in S_{n\cdot m}}$ can be identified with the multiplet

${\displaystyle (\pi _{x};\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)})\in S_{n}\times S_{m}^{n},}$

which will be written in twisted form in the next section.

The permutations ${\displaystyle \gamma _{x}}$, which arose as second components of the monomial representation

${\displaystyle {\begin{cases}G\to K\wr S_{n\cdot m}\\x\mapsto (k_{x}(1,1),\ldots ,k_{x}(n,m);\gamma _{x})\end{cases}}}$

in the previous section, are of a very special kind. They belong to the stabilizer of the natural equipartition of the set ${\displaystyle [1,n]\times [1,m]}$ into the ${\displaystyle n}$ rows of the corresponding matrix (rectangular array). Using the peculiarities of the composition of Artin transfers in the previous section, we show that this stabilizer is isomorphic to the wreath product ${\displaystyle S_{m}\wr S_{n}}$ of the symmetric groups ${\displaystyle S_{m}}$ and ${\displaystyle S_{n}}$ with respect to the set ${\displaystyle \{1,\ldots ,n\}}$, whose underlying set ${\displaystyle S_{m}^{n}\times S_{n}}$ is endowed with the following law of composition:

${\displaystyle {\begin{aligned}(10)\quad \forall x,z\in G:\qquad \gamma _{x}\cdot \gamma _{z}&=(\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})\cdot (\sigma _{u_{z}(1)},\ldots ,\sigma _{u_{z}(n)};\pi _{z})\\&=(\sigma _{u_{x}(\pi _{z}(1))}\circ \sigma _{u_{z}(1)},\ldots ,\sigma _{u_{x}(\pi _{z}(n))}\circ \sigma _{u_{z}(n)};\pi _{x}\circ \pi _{z})\\&=(\sigma _{u_{xz}(1)},\ldots ,\sigma _{u_{xz}(n)};\pi _{xz})\\&=\gamma _{xz}\end{aligned}}}$

This law reminds of the chain rule ${\displaystyle D(g\circ f)(x)=D(g)(f(x))\circ D(f)(x)}$ for the Fréchet derivative in ${\displaystyle x\in E}$ of the compositum of differentiable functions ${\displaystyle f:E\to F}$ and ${\displaystyle g:F\to G}$ between complete normed spaces.

The above considerations establish a third representation, the stabilizer representation,

${\displaystyle {\begin{cases}G\to S_{m}\wr S_{n}\\x\mapsto (\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})\end{cases}}}$

of the group ${\displaystyle G}$ in the wreath product ${\displaystyle S_{m}\wr S_{n}}$, similar to the permutation representation and the monomial representation. As opposed to the latter, the stabilizer representation cannot be injective, in general. For instance, certainly not, if ${\displaystyle G}$ is infinite. Formula (10) proves the following statement.

Theorem. The stabilizer representation

${\displaystyle {\begin{cases}G\to S_{m}\wr S_{n}\\x\mapsto \gamma _{x}=(\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})\end{cases}}}$

of the group ${\displaystyle G}$ in the wreath product ${\displaystyle S_{m}\wr S_{n}}$ of symmetric groups is a group homomorphism.

Let ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of a subgroup ${\displaystyle H}$ of finite index ${\displaystyle n}$ in a group ${\displaystyle G}$ and ${\displaystyle x\mapsto \pi _{x}}$ be its associated permutation representation.

Theorem.^{[1][3][4][5][8][9]} Suppose the permutation ${\displaystyle \pi _{x}}$ decomposes into pairwise disjoint (and thus commuting) cycles ${\displaystyle \zeta _{1},\ldots ,\zeta _{t}\in S_{n}}$ of lengths ${\displaystyle f_{1},\ldots f_{t},}$ which is unique up to the ordering of the cycles. More explicitly, suppose

${\displaystyle (11)\quad \left(g_{j}H,g_{\zeta _{j}(j)}H,g_{\zeta _{j}^{2}(j)}H,\ldots ,g_{\zeta _{j}^{f_{j}-1}(j)}H\right)=\left(g_{j}H,xg_{j}H,x^{2}g_{j}H,\ldots ,x^{f_{j}-1}g_{j}H\right),}$

for ${\displaystyle 1\leq j\leq t}$, and ${\displaystyle f_{1}+\cdots +f_{t}=n.}$ Then the image of ${\displaystyle x\in G}$ under the Artin transfer is given by

${\displaystyle (12)\quad T_{G,H}(x)=\prod _{j=1}^{t}g_{j}^{-1}x^{f_{j}}g_{j}\cdot H'.}$

The cycle decomposition corresponds to a ${\displaystyle (\langle x\rangle ,H)}$ double coset decomposition of ${\displaystyle G}$:

${\displaystyle G=\bigsqcup _{j=1}^{t}\langle x\rangle g_{j}H}$

It was this cycle decomposition form of the transfer homomorphism which was given by E. Artin in his original 1929 paper.^[3]

Let ${\displaystyle H}$ be a normal subgroup of finite index ${\displaystyle n}$ in a group ${\displaystyle G}$. Then we have ${\displaystyle xH=Hx}$, for all ${\displaystyle x\in G}$, and there exists the quotient group ${\displaystyle G/H}$ of order ${\displaystyle n}$. For an element ${\displaystyle x\in G}$, we let ${\displaystyle f:=\mathrm {ord} (xH)}$ denote the order of the coset ${\displaystyle xH}$ in ${\displaystyle G/H}$