## Burnside’s normal complement theorem (2)

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

As part (1) we’ll assume that $G$ is a group and $H$ is an abelian subgroup of $G$ with $[G:H]=n < \infty.$ We will also fix a transversal $T=\{g_1, \cdots , g_n \}$ of $H$ in $G.$ Let $\lambda$ be the group homomorphism introduced in the theorem we proved in part (1).

Theorem. There exist positive integers $m, k_1, \cdots , k_m$ and elements $x_1, \cdots , x_m \in G$ such that $\sum_{j=1}^m k_j = n$ and $\lambda(g)= \prod_{j=1}^m x_j^{-1}g^{k_j} x_j,$ for all $g \in G.$

Proof. Let $g \in G.$ By Lemma 1 in part (1), there exists $\alpha \in S_n$ and $h_i \in H$ such that

$gg_i = g_{\alpha(i)} h_i \ \ \ \ \ \ \ \ \ \ \ (1)$

for all $i.$ Now let $\alpha = \alpha_1 \alpha_2 \cdots \alpha_m$ be the decomposition of $\alpha$ into disjoint cycles. Let

$\alpha_j = (t_{1,j} \ t_{2,j} \ \cdots t_{k_j, j}),$

for all $1 \leq j \leq m.$ Clearly $\sum_{j=1}^m k_j = n.$ Now by (1) we have

$g g_{t_{i,j}}=g_{t_{i+1,j}}h_{t_{i,j}}, \ \ \ \ \ \ \ \ \ \ \ (2)$

if $i < k_j$ and

$gg_{t_{k_j,j}}=g_{t_{1,j}}h_{t_{k_j,j}}. \ \ \ \ \ \ \ \ \ \ \ (3).$

If in (2) we let $i=1$ and multiply from the left by $g,$ we will get

$g^2g_{t_{1,j}}=gg_{t_{2,j}}h_{t_{1,j}}. \ \ \ \ \ \ \ \ \ \ (4)$

But if in (2) we let $i=2,$ we will get

$gg_{t_{2,j}}=g_{t_{3,j}}h_{t_{2,j}}. \ \ \ \ \ \ \ \ \ \ (5)$

So if we replace $g g_{t_{2,j}}$ in the right side of (4) with what I got in (5), we will get $g^2g_{t_{1,j}}=g_{t_{3,j}}h_{t_{2,j}}h_{t_{1,j}}.$ If we repeat this process, we will get $g^3 g_{t_{1,j}} = g_{t_{4,j}}h_{t_{3,j}}h_{t_{2,j}}h_{t_{1,j}}.$ If we continue in this manner, paying attention to the last step that we’ll have to use (3), we will eventually get

$g^{k_j} g_{t_{1,j}} = g_{t_{1,j}} \prod_{i=1}^{k_j} h_{t_{i,j}}. \ \ \ \ \ \ \ \ \ \ (6)$

Thus if we let $x_j = g_{t_{i,j}},$ then (6) gives us

$x_j^{-1} g^{k_j} x_j = \prod_{i=1}^{k_j} h_{t_{i,j}}. \ \ \ \ \ \ \ \ \ \ \ \ \ (7).$

Note that since $\alpha_1 \alpha_2 \cdots \alpha_m$ is the cycle decomposition of $\alpha,$ we have $\{1,2, \cdots , n \} = \bigcup_{i,j} t_{i,j}$ and so $\{h_{t_{i,j}}: \ 1 \leq i \leq k_j, \ 1 \leq j \leq m \} = \{h_1, h_2, \cdots , h_n \}.$ Thus the definition of $\lambda(g)$ and (7) gives us

$\lambda(g)=\prod_{i=1}^n h_i = \prod_{j=1}^m \prod_{i=1}^{k_j} h_{t_{i,j}}=\prod_{j=1}^m x_j^{-1} g^{k_j} x_j. \ \Box$