Focal subgroup theorem

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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.

Related facts

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

Journal 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)