## 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 $n_p(G) \geq n_p(G/N)$.

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 $gH$.