Focal subgroup theorem

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow subgroup) must also satisfy the second subgroup property (i.e., subgroup whose focal subgroup equals its intersection with the commutator subgroup)
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Statement

Let $P$ be a $p$-Sylow subgroup of a finite group $G$ and let:

$P_0=⟨x{y}^{-1}\mid x,y\in P,\mathrm{\exists }g\in G,gx{g}^{-1}=y⟩$.

In other words, $P_0$ is the focal subgroup of $P$ in $G$.

Then:

$P\cap G^{\prime }=P_0$

In other words, $P$ is a subgroup whose focal subgroup equals its intersection with the commutator subgroup.

Related facts

• Analogue of focal subgroup theorem for Hall subgroups

Facts used

For the proof using linear representation theory

1. Brauer's induction theorem

For the proof using the transfer homomorphism

1. Product decomposition for element in terms of transfer homomorphism

Proof

Given: A finite group $G$, a $p$-Sylow subgroup $P$. $P_0$ is the focal subgroup of $P$, defined by:

$P_0=⟨x{y}^{-1}\mid x,y\in P,\mathrm{\exists }g\in G,gx{g}^{-1}=y⟩$.

Further, we have:

$G^{\prime }=[G,G]=⟨x{y}^{-1}\mid x,y\in G,\mathrm{\exists }g\in G,gx{g}^{-1}=y⟩$.

To prove: $P_0=P\cap G^{\prime }$.

Initial observations

First, note that $P^{\prime }\subseteq P_0\subseteq P\cap G^{\prime }$. The first inclusion is because every commutator of elements in $P$ is contained inside the generating set for $P_0$, and the second inclusion is because every element in the given generating set for $P_0$ is both in $P$ and in $G_0$.

Since $P_0$ contains $P^{\prime }$, $P/P_0$ is an abelian group.

Proof using linear representation theory

Claim: Given any linear character on $P/P_0$, there exists a linear character on $G$ extending it.

Proof:

1. Let $p_1,p_2,\dots ,p_m$ be the distinct prime divisors of $\left|G\right|$ with $p=p_1$. For each $p_i$, pick a $p_i$-Sylow subgroup $P_i$ such that $P_1=P$.
2. Any $g\in G$ can be expressed as a product $g=g_1{g}_2\dots g_m$ where each $g_i$ is conjugate to some $h_i\in P_i$ and the $g_i$ commute pairwise. If $g_1$ is also conjugate to ${h_{1^{\prime }}}^{\prime }\in P_1$ then $h_1{h}_1{\text{'}}^{-1}\in P_0$.
3. Let $\lambda$ be a linear character of $P$ whose kernel contains $P_0$. We claim that the function $\theta (g)=\lambda (h_1)$, where $h_1$ is as described above, is well-defined: The reason is that if $h_1$ and ${h_{1^{\prime }}}^{\prime }$ are two possibilities, then $\lambda (h_1{h}_1{\text{'}}^{-1})=1$ and hence $\lambda (h_1)=\lambda ({h_{1^{\prime }}}^{\prime })$.
4. $\theta$ is a linear character of $G$: Clearly, $\theta$ is a class function on $G$. Then by fact (1), we conclude that $\theta$ is a linear character of $G$. Further, $\theta$ extends $\lambda$, viz., the restriction of $\theta$ to $P$ is $\lambda$.

How the result follows from the claim: We now show that $P_0=P\cap G^{\prime }$. Suppose $P_0$ is properly contained in $P\cap G^{\prime }$. Then, we can choose a linear character $\lambda$ of the abelian group $P/P_0$ that is nontrivial on some element of $G^{\prime }$. However, such a character cannot be extended to a linear character of the whole group, since any linear character of the whole group must be trivial on all elements of $G^{\prime }$. Thus, $P_0$ cannot be properly contained in $P\cap G^{\prime }$.

Proof using the transfer homomorphism

We have $P_0\le P\le G$, with $P_0$ normal in $P$ and $P/P_0$ abelian. In particular, we can construct the transfer homomorphism $\tau :G\to P/P_0$. Let $\phi :P\to P/P_0$ be the quotient map.

Claim: The restriction of $\tau$ to $P$ is surjective to $P/P_0$.

Proof: For any $x\in P$, we have, by fact (1), a collection of elements $x_1,x_2,\dots ,x_t\in G$ and nonnegative integers $r_1,r_2,\dots ,r_t$ such that $\sum r_i=[G:P]$, and we have:

$x_i{x}^{r_i}{x}_i^{-1}\in H,\tau (x)=\prod _{i=1}^t{x}_i{x}^{r_i}{x}_i^{-1}mod{P}_0$.

Since we chose $x\in P$, and $P/P_0$ is abelian, we can rearrange terms to obtain:

$\tau (x)=\prod _{i=1}^t{x}^{r_i}\prod _{i=1}^t{x}^{-r_i}{x}_i{x}^{r_i}{x}_i^{-1}mod{P}_0$.

Every term in the second product is of the form $x^{-r_i}(x_i{x}^{r_i}{x}_i^{-1})$, which is of the form $a{b}^{-1}$ with $a,b\in P$ conjugate in $G$ (here $a=x^{-r_i}$). In particular, every term in the second product is in $P_0$, and we obtain:

$\tau (x)=\prod _{i=1}^t{x}^{r_i}mod{P}_0$,

which, since $\sum r_i=[G:P]$, reduces to:

$\tau (x)=x^{[G:P]}mod{P}_0$.

Note now that since $P$ is $p$-Sylow, $[G:P]$ is relatively prime to $p$, and the map $x\mapsto x^{[G:P]}$ is a bijection from $P/P_0$ to itself. Thus, $\tau$, restricted to $P$, surjects to $P/P_0$.

Proof using the claim: Let $K$ be the kernel of $\tau :G\to P/P_0$.

Since $K$ is a kernel of a map to an abelian group, $[G,G]\le K$. Thus, $P_0\le [G,G]\cap P\le K\cap P$. The restriction of $\tau$ to $P$ is a bijective map from $P/(P\cap K)$ to $P/P_0$ (surjective by the claim, and injective because we've quotiented by the kernel). Thus, the size of $P\cap K$ equals the size of $P_0$ forcing $P_0=[G,G]\cap P=K\cap P$. In particular, $P_0=[G,G]\cap P$.

References

Journal references

• Focal series in finite groups by Donald Gordon Higman, , Volume 5, Page 477 - 497(Year 1953): ^{More info}

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 250, Theorem 3.4, Section 7.3 (Transfer and the focal subgroup)