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)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about Sylow subgroup|Get more facts about subgroup whose focal subgroup equals its intersection with the commutator subgroup

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:

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

and:

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

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

Initial observations

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$P^{\prime }\subseteq P_0\subseteq P\cap G^{\prime }$				[SHOW MORE] 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 $G^{\prime }$.
2	$P_0$ is normal in $P$ and $P/P_0$ is abelian.			Step (1)	This follows from the part of Step (1) that asserts that $P^{\prime }\le P_0$.
3	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 $P_1=P$.
4	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.	Fact (2)		Step (3)	[SHOW MORE] Consider the cyclic group $⟨g⟩$. This is a direct product of its $p_i$-Sylow subgroups, so $g$ can be expressed as a product of commuting elements $g_i$, one from each Sylow subgroup. $g_i$ is an element of order a power of $p_i$, so by Fact (2), $g_i$ is conjugate to an element $h_i\in P_i$.
5	If $g_1$ (continuing notation from Step (4), though it does not really matter) is conjugate to $h_1\in P_1$ and also conjugate to ${h_{1^{\prime }}}^{\prime }\in P_1$ then $h_1{h}_1{\text{'}}^{-1}\in P_0$.				[SHOW MORE] Because being conjugate is an equivalence relation, and the elements $h_1$ and ${h_{1^{\prime }}}^{\prime }$ are both conjugate to $g_1$, they are conjugate to each other in $G$. Hence, by the definition of $P_0$, $h_1{h}_1{\text{'}}^{-1}$ is in the generating set for $P_0$, and hence is in $P_0$.
6	Let $\lambda$ be a linear character (i.e., the character of a one-dimensional representation) of $P$ whose kernel contains $P_0$. The function $\theta (g)=\lambda (h_1)$, where $h_1$ is as described in Step (4), is well-defined			Steps (4), (5)	[SHOW MORE] The reason is that if $h_1$ and ${h_{1^{\prime }}}^{\prime }$ are two possibilities, then by Step (5), $h_1{h}_1^{-1}\in P_0$, so $\lambda (h_1{h}_1{\text{'}}^{-1})=1$ and hence $\lambda (h_1)=\lambda ({h_{1^{\prime }}}^{\prime })$ on account of $\lambda$ being multiplicative since it's a linear character.
7	Continuing with the notation of Step (6), $\theta$ is a class function on $G$.				[SHOW MORE] If $g,g^{\prime }$ are conjugate by $y$, then $g{\text{'}}_1$ is the conjugate of $g^{\prime }$ by $y$. Hence, the element $h_1$ of $P_1$ that we choose as conjugate to $g_1$ is also conjugate to $g{\text{'}}_1$, hence the definition of $\theta$ gives the same value to $\theta (g)$ and $\theta (g^{\prime })$.
8	Continuing with the notation of Step (6), $\theta$ is a linear character of $G$. In other words, $\theta (g{g}^{\prime })=\theta (g)\theta (g^{\prime })$ for all $g,g^{\prime }\in G$.				This critical step is unclear
9	For any linear character $\lambda$ of $P$ whose kernel contains $P_0$, there is a linear character $\theta$ of $G$ that extends $\lambda$, i.e., the restriction of $\theta$ to $P$ is $\lambda$.			Steps (6)-(8)	Step-combination direct, plus: [SHOW MORE] We note that if $g\in P_1=P$ to begin with, then $g=g_1=h_1$, so $\theta (g)=\lambda (h_1)=\lambda (g)$. Thus, the restriction of $\theta$ to $P$ is $\lambda$.
10	$P\cap G^{\prime }\le P_0$.				[SHOW MORE] Suppose not. Then, we can choose a linear character $\lambda$ of $P$ with kernel containing $P_0$ that is nontrivial on an element of $P\cap G^{\prime }$ that is outside $P_0$. 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 $G^{\prime }$. On the other hand, Step (9) shows that such a character can be extended to the whole group. Our assumption thus leads to a contradiction.
11	$P_0=P\cap G^{\prime }$.			Steps (1), (10)	Step-combination direct.

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)