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

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