Let be a finite group and suppose is normal. Prove that .

We begin with a lemma.

Lemma: Every Sylow -subgroup of has the form for some Sylow -subgroup of . Proof: Let be a Sylow -subgroup of . In this previous exercise, we saw that is a Sylow -subgroup of . Now let be any Sylow -subgroup; then for some by Sylow’s Theorem, and is a Sylow -subgroup of .

Now to the main result.

Note that the mapping given by is a surjection of finite sets. Thus .

## Comments

can you explain the operation gN PN/N g^-1N = gPg^-1N/N

Sure. There is a bit of abuse of notation happening here. We let be a Sylow subgroup, which by Sylow’s theorem is conjugate to . Since the elements of are just cosets, we have for some element . Using the natural homomorphism and since is normal, we have as sets.

I should probably rewrite this…one subtlety that I didn’t understand as well when I wrote a lot of these solutions is that the ‘mod ‘ notation is really an abuse of notation itself. is a partition of by an equivalence relation induced by . So with , the ‘quotient’ is really a set of equivalence classes, as is , even though is not contained in .

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