Why $n_i=j$ implies that there is a subgroup $H$ of $G$ of index j? 
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 ?
 A: 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 $H \in O(P)$.
A: 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 $G$.
