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Normal Subgroups, Centralizers and Semi-Direct Products

1. I f A and B are normal subgroups of G such that G/A and G/B are abelian, prove that G/(A intersect B) is abelian

2. Let H and K be groups, let f be a homomorphism from K into Aut(H) and as usual identify H and K as subgroups of G= H x_f K( x_f denotes product of H and K under f).
Prove that C_K(H)= Ker(f)
ps. C_K(H) is centralizer

keywords: semi direct

Solution Preview

1. Proof:
We use A^B to denote "A intersect B".
We consider two element g(A^B) and h(A^B) in G/(A^B), we want to show
that g(A^B) * h(A^B) = gh(A^B) = h(A^B) * g(A^B) = hg(A^B)
That is equivalent to show that g^(-1)h^(-1)gh belongs to A^B.
Since G/A is abelian, then gA and hA are commutative, so we have
gA * hA = ghA = hA * gA = hgA.
This implies that ...

Solution Summary

Normal Subgroups, Centralizers and Semi-Direct Products are investigated.

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