## Burnside’s normal complement theorem (3)

Posted: January 20, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $G=PQ. \ \Box$