- #1
Bachelier
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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.
I'm confused about this part.
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)]
The normalizer of H is a subgroup containing H, so since H has index 3, either NG (H) = H or NG (H) = G.
I'm confused about this part.
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)]