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Justabeginner
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Homework Statement
Two subgroups of G, H and K are conjugate if an element a in G exists such that aHa^-1= {aha^-1|elements h in H}= K
Prove that if G is finite, then the number of subgroups conjugate to H equals |G|/|A|.
Homework Equations
A={elements a in G|aHa^-1=H}
The Attempt at a Solution
Stabilizer of H also known as NG(H) is {a: aHa^-1 = H}.
By Lagrange's Theorem, number of left cosets in H = |G|/|H|.
NG (H) is a normalizer and NG (H) = {elements g in G| gHg^-1 = H}.
A is normalizer and A= NG (H) because A={elements a in G|aHa^-1=H}.
Therefore, no. of subgroups conjugate to H, is |G|/|A|.
This argument is terribly weak and I have no idea how to formalize the proof. I want to just use the concept of subgroup/group properties not normalizers/stabilizers/centralizers because I clearly do not understand the latter well enough to do so.
I appreciate any help. Thank you.