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.(adsbygoogle = window.adsbygoogle || []).push({});

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)]

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

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