Calculating the number of Sylow $5$- and $7$-subgroups in a group of order $105$

Question

I was reading the solution to a problem, but there's a part I really don't understand.

So basically, you have to calculate the $5$-Sylow and the $7$-Sylow subgroups of a group G of order $105$, and by cardinality arguments, you find out that at least one of them must be normal. Cool. Also, you can show that $PQ$ is abelian, where $P$ is a $5$-Sylow subgroup and $Q$ is a $7$-Sylow subgroup. But now there's a way to show that both $P$ and $Q$ are normal in G, and it's the following:

Both $P$ and $Q$ are normalized by the whole of the abelian subgroup $PQ$, so the number of $5$-Sylow and $7$-Sylow subgroups is less than $|G|/|PQ| = 3$.

And I really don't understand why this is true. Can someone help me?

Answer 1

If $P$ is one particular Sylow $p$-subgroup of $G$, then the other Sylow $p$-subgroups of $G$ are precisely the conjugates of $P$. Thus $$ \mathcal P = \{ gPg^{-1} : g \in G \}$$ is the set of all Sylow $p$-subgroups of $G$.

Notice that there is a group action of $G$ on $\mathcal P$: $G$ acts on $\mathcal P$ by conjugation. If $\phi: G \times \mathcal P \to \mathcal P$ is this group action, then $ \phi(g, Q) = gHg^{-1}$ for all $g \in G$ and $H \in \mathcal P$.

Now observe that:

• $\text{Orb}(P)$, the orbit of $P$, contains the whole of $\mathcal P$. In other words, $\text{Orb}(P)$ is the set of all the Sylow $p$-subgroups of $G$.
• $\text{Stab}(P)$, the stabiliser of $P$, is the same thing as the normaliser of $P$ in $G$. In other words, $\text{Stab}(P)$ is equal to $N_G(P): = \{ g \in G : gPg^{-1} = P \}$.

Thus $$ |\mathcal P | = |\text{Orb}(P)| = \frac{|G|}{|\text{Stab}(P)|} = \frac{|G|}{|N_G(P)|}.$$

Since $P$ is a subgroup of $PQ$ and $PQ$ is an abelian group, $P$ must be a normal subgroup of $PQ$. In other words, $gPg^{-1} = P$ for all $g \in PQ$, i.e. $P$ is "normalised" by every $g \in PQ$.

Thus $PQ$ is a subgroup of $N_G(P)$, so $$ |\mathcal P | = \frac{|G|}{|N_G(P)|} \leq \frac{|G|}{|PQ|},$$ as claimed.