Burnside’s normal complement theorem (3)

Posted: January 20, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , , , ,

Notation. For a group G and a subgroup H of G, the centralizer and the normalizer of H in G are denoted by C(H) and N(H) respectively.

Lemma. Let G be a finite group and let P be a Sylow subgroup of G. Suppose that a,b \in C(P) and b=x^{-1}ax for some x \in G. There exists y \in N(P) such that b=y^{-1}ay.

Proof. Let C(b)=\{g \in G: \ gb=bg \}, the centralizer of b in G. Clearly P \subseteq C(b) because b \in C(P) and so b commutes with every element of P. Also if c \in P, then

(x^{-1}cx)b = (x^{-1}cx)(x^{-1}ax)=x^{-1}cax=x^{-1}acx=b(x^{-1}cx).

Thus x^{-1}Px \subseteq C(b). So both P and x^{-1}Px are Sylow subgroups of C(b) and so, since |P|=|x^{-1}Px|, there exists z \in C(b) such that P= z^{-1}x^{-1}Pxz. Then clearly y=xz \in N(P) and, since z \in C(b), we have y^{-1}ay = z^{-1}x^{-1}axz=z^{-1}bz=b. \ \Box

Burnside’s Normal Complement Theorem. (Burnside, 1900) Let G be a finite group and let P be a Sylow subgroup of G. If C(P)=N(P), then there exists a normal subgroup Q of G such that P \cap Q = \{1\} and G=PQ.

The normal subgroup Q in the theorem is called the normal complement of P.

Proof. Clearly P is abelian because P \subseteq N(P) = C(P), Let |P|=p^r and [G:P]=n. Note that, since P is a Sylow subgroup, \gcd(n,p^r)=1 and hence there exist integers u,v such that

nu+p^rv=1. \ \ \ \ \ \ \ \ \ (1).

Now let g \in P. The theorem in part (2) gives us elements x_1, \cdots , x_m \in G and integers k_1, \cdots , k_m such that \sum_{j=1}^m k_j = n and \lambda(g)= \prod_{j=1}^m x_j^{-1}g^{k_j}x_j, where \lambda is the group homomorphism defined in the theorem in part (1). Note that, by the equation (7) in the proof of the theorem in part (2), x_j^{-1}g^{k_j} x_j \in P \subseteq C(P) for all j. Thus, by the Lemma, for every j there exists some y_j \in N(P) such that x_j^{-1}g^{k_j}x_j=y_j^{-1}g^{k_j}y_j. Hence x_j^{-1}g^{k_j}x_j = g^{k_j}, for all j, because y_j \in N(P)=C(P). Thus

\lambda(g) = \prod_{j=1}^m x_j^{-1}g^{k_j}x_j = \prod_{j=1}^m g^{k_j} = g^{\sum_{j=1}^m k_j}=g^n. \ \ \ \ \ \ \ \ \ \ (2)

Now,  (1) and (2) give us

\lambda(g^u)=g^{nu}=g^{1-p^rv}=g, \ \ \ \ \ \ \ \ \ \ \ (3)

because |P|=p^r and so g^{p^r}=1. So (3) shows that \lambda is onto. Let \ker \lambda = Q.  Then

G/Q \cong P. \ \ \ \ \ \ \ \ \ \ \ (4)

We can now easily prove that Q is the required subgroup in the theorem:

1) Obviously Q is normal because Q is the kernel of the group homomorphism \lambda. To see why P \cap Q = \{1\}, let g \in P \cap Q. Then g^{p^r}=1 and g^n = \lambda(g)=1. But then using (1) we will have g = g^{nu + p^rv}=1.

2) To see why G=PQ, we first note that |PQ|=|P| \cdot |Q|, because |P \cap Q|=1 as we just showed. We also have |P| \cdot |Q|=|G|, by (4). Thus |G|=|PQ| and hence G=PQ. \ \Box

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s