Why $n_i=j$ implies that there is a subgroup $H$ of $G$ of index j?

Question

Why $n_i=j$ implies that there is a subgroup $H$ of $G$ of index j ?

If I denote with $n_i$ the number of i-Sylow subgroups

For example if $|G|=180=2^2\cdot3^2\cdot 5$

then if I assume that $n_5=6$ then $G$ has a subgroup of index $6$, why is this true ?

Answer 1

Let $G$ operate by conjugation on its Sylow $p$-groups and let $P$ be one of those Sylow-groups. By the Sylow theorems, all Sylow $p$-groups are conjugate and hence, $O(P)$ (the orbit of $P$ under conjugation) contains all the Sylow $p$-groups. If $n_p$ is the number of Sylow $p$-groups, we have $$n_p=|O(P)|=[G:N_G(P)]$$

So, the normalizer of any Sylow $p$-group has index $n_p$ in $G$.

$\textbf{Edit:}$ Let me show that we have indeed $|O(P)|=n_p$. It suffices to show the following:

$\phantom{aaaaaaaaaaaaaaaaaaaaaaaaa}H \in O(P)\phantom{aaa}$ iff $\phantom{aaa}H$ is a Sylow $p$-group

Assume that $H \in O(P)$. Then $H$ is conjugate to $P$ which means that we find an element $g \in G$ with $gPg^{-1}=H$. But then, $P$ and $H$ are isomorphic (the corresponding isomorphism from $P$ to $H$ is given by $a \mapsto gag^{-1}$). Therefore, $|H|=|P|$ and this means that $H$ is a Sylow $p$-group.

For the other direction, assume $H$ is a Sylow $p$-group. Then, by the theorem of Sylow, $H$ and $P$ are conjugate which means that $H \in O(P)$.

Answer 2

Consider the action of $G$ on $Syl_{i}\left(G\right)$ by conjugation. This action is transitive thanks to second Sylow theorem. So, by the orbit-stabilizer theorem, we have: $$n_{i}=\left|Syl_{i}\left(G\right)\right|=\frac{\left|G\right|}{\left|Stab_{G}\left(P\right)\right|}=\left[G:N_{G}\left(P\right)\right] $$ $P$ is $i$-Sylow subgroup of $G$.