Alternate way to prove that $p_i$- sylow subgroup is unique in nilpotent group

Question

Let $G$ is nilpotent group, $|G| = n = p_1^{\alpha_1} \times \cdots p_l^{\alpha_l} $ be the prime factorization of $n$ then in order to show that $p_i$ sylow subgroup is unique I just need to show that

$\text{number of elements of order } p_i^1 + \text{number of elements of order } p_i^2 + \text{number of elements of order } p_i^3 + \cdots \text{number of elements of order } p_i^m = p_i^{\alpha_i}$

Question : Is there other way to prove that $G$ has $p_i$- sylow subgroup unique?

Answer 1

It's well-known that nilpotent subgroups satisfy the Normalizer Property, namely if $H < G$, where $G$ is nilpotent then we have that $H$ is a proper subgroup of $N[H]$, the normalizer of $H$.

Now let $H_1$ be a Sylow $p_1$-subgroup of $G$. If $H_1$ isn't a proper subgroup, then the claim is trivial, as $G$ is a $p$-grpup. So let $H_1$ be a proper subgroup of $G$. Then it's not hard to prove that $N[N[H_1]] = N[H_1]$. But if $N[H_1]$ is a proper subgroup, by the Normalize Property this is impossible. Therefore $N[H_1] = G$ and hence $H_1$ is normal in $G$. And we know that normal Sylow subgroups are unique of the given order.

Answer 2

Suppose $G$ is nilpotent, that is, It has a central series

$$1=Z_0<Z_1<... Z_{n-1}<Z_n=G$$

Let's proceed induction on central length $n$,

if $n=1$, $G$ is abelain and claim is true. Now assume the claim is true for groups whose lengt at most $n-1$.

Let $P\in Syl_p(G)$, then $PZ_1/Z_1\lhd G/Z_1$ by induction. Thus $PZ_1\lhd G$. Notice that $P\lhd PZ_1$ as $Z_1$ is central. $P$ is unique Sylow subgroup of $PZ_1$ and $P^g\leq (PZ_1)^g=PZ_1$. It follows $P=P^g$ and hence $P\lhd G$.