An upper bound on the number of Sylow subgroups of a finite quotient

Let G be a finite group and suppose N \leq G is normal. Prove that n_p(G/N) \leq n_p(G).

We begin with a lemma.

Lemma: Every Sylow p-subgroup of G/N has the form PN/N for some Sylow p-subgroup P of G. Proof: Let P be a Sylow p-subgroup of G. In this previous exercise, we saw that PN/N is a Sylow p-subgroup of G/N. Now let Q \leq G/N be any Sylow p-subgroup; then Q = (gN)(PN/N)(g^{-1}N) = (gPg^{-1})N/N for some g \in G by Sylow’s Theorem, and gPg^{-1} is a Sylow p-subgroup of G. \square

Now to the main result.

Note that the mapping \Phi : \mathsf{Syl}_p(G) \rightarrow \mathsf{Syl}_p(G/N) given by P \mapsto PN/N is a surjection of finite sets. Thus n_p(G) \geq n_p(G/N).

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  • neal  On August 6, 2011 at 9:05 pm

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

    • nbloomf  On September 6, 2011 at 10:15 am

      Sure. There is a bit of abuse of notation happening here. We let Q be a Sylow subgroup, which by Sylow’s theorem is conjugate to PN/N. Since the elements of G/N are just cosets, we have Q = (gN) (PN/N) (g^{-1}N) for some element g \in G. Using the natural homomorphism G \rightarrow G/N and since N is normal, we have (gN) (PN/N) (g^{-1}N) = (gPg^{-1}N)/N 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 N‘ notation is really an abuse of notation itself. G/N is a partition of G by an equivalence relation induced by N. So with N \leq H, the ‘quotient’ H/N is really a set of equivalence classes, as is gH/N, even though N is not contained in gH.


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