# 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:

$P_0 = \langle xy^{-1} \mid x,y \in P, \exists g \in G, gxg^{-1} = y \rangle$.

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

Then:

$P \cap G' = P_0$

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

## 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:

$P_0 = \langle xy^{-1} \mid x,y \in P, \exists g \in G, gxg^{-1} = y \rangle$.

Further, we have:

$G' = [G,G] = \langle xy^{-1} \mid x,y \in G, \exists g \in G, gxg^{-1} = y \rangle$.

and:

$P' = [P,P] = \langle xy^{-1} \mid x,y \in P, \exists g \in P, gxg^{-1} = y \rangle$.

To prove: $P_0 = P \cap G'$.

### Initial observations

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $P' \subseteq P_0 \subseteq P \cap G'$ [SHOW MORE]
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 $P' \le P_0$.
3 Let $p_1, p_2, \ldots, p_m$ be the distinct prime divisors of $|G|$ 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 $g = g_1g_2\ldots g_m$ 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]
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 $h_1' \in P_1$ then $h_1h_1'^{-1} \in P_0$. [SHOW MORE]
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]
7 Continuing with the notation of Step (6), $\theta$ is a class function on $G$. [SHOW MORE]
8 Continuing with the notation of Step (6), $\theta$ is a linear character of $G$. In other words, $\theta(gg') = \theta(g)\theta(g')$ for all $g,g' \in G$. 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]
10 $P \cap G' \le P_0$. [SHOW MORE]
11 $P_0 = P \cap G'$. 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 $\varphi:P \to P/P_0$ 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:

$x_i x^{r_i}x_i^{-1} \in H, \qquad \tau(x) = \prod_{i=1}^t x_ix^{r_i}x_i^{-1} \mod P_0$.

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

$\tau(x) = \prod_{i=1}^t x^{r_i} \prod_{i=1}^t x^{-r_i}x_ix^{r_i}x_i^{-1} \mod P_0$.

Every term in the second product is of the form $x^{-r_i}(x_ix^{r_i}x_i^{-1})$, which is of the form $ab^{-1}$ 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

### Textbook references

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