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

[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 N_G (H) = H or N_G (H) = G.

I'm confused about this part.

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

 

[micromass]

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

 

[Deveno]

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