# Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b.

1. Dec 9, 2011

### Bachelier

Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b. Let H = { 1, a3, b, a3b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H.

The normalizer of H is a subgroup containing H, so since H has index 3, either NG (H) = H or NG (H) = G.

By Lagrange, |G|=|H|[G], since normalizer must contain H, then NG(H)≥H. But then its order can be 6. I'm missing something here about [G:NG(H)]

2. Dec 9, 2011

### micromass

Re: Normalizer

I honestly don't understand your question
Everything you say is true. So what's confusing you?

3. Dec 10, 2011

### Deveno

Re: Normalizer

no the order cannot be 6. NG(H) contains H as a subgroup. H is of order 4, and 4 does not divide 6.