Focal subgroup theorem

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.

Related facts

 * Analogue of focal subgroup theorem for Hall subgroups

For the proof using linear representation theory

 * 1) uses::Brauer's induction theorem

For the proof using the transfer homomorphism

 * 1) uses::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: $$P_0 = P \cap G'$$.

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, $$P_0 = [G,G] \cap P$$.

Textbook references

 * , Page 250, Theorem 3.4, Section 7.3 (Transfer and the focal subgroup)