I've been struggling with this exercise (Exercise 8.2.28, Introduction to Abstract Algebra by Nicholson) for a few hours, and I haven't made much progress.
Let $G$ be a group of order $p^n$ and let $H_1, \dots ,H_m$ be the distinct subgroups of $G$ of index $p$. If $N = H_1 \cap \cdots \cap H_m$, show that $N \triangleleft G$ and that $x^p = 1$ for every coset $x$ in $G/N$.
From an earlier exercise, either $H_i \triangleleft G$ or $H_i = N(H_i)$ where $N(H_i)$ is the normalizer of $H_i$ for each $i$.
Furthermore, by Lagrange's Theorem, these subgroups are the ones of order $p^{n-1}$.
Let $G$ be a finite $p$-group of order $p^n$. Then there exists a series $$ G = G_0 \supset G_1 \supset \cdots \supset G_n = \{ 1 \} $$ of subgroups of $G$ such that $G_i \triangleleft G$, $|G_i| = p^{n-i}$, and $|G_i / G_{i+1}| = p$ for all $i$.
I have a strong feeling that this is somehow related but I've been stuck for hours on how to use it. I have no idea where to go from here and I've made so little progress. Can someone provide me a hint rather than a full answer? I still want to be challenged.
To show that $N$ is normal, note that if $H$ is of index $p$, then the index of $H$ is the smallest prime dividing $|G|$, so $H$ is normal. Thus, $N$ is the intersection of normal subgroups, hence is normal.
Alternatively, if $H$ has index $p$ and $x\in G$, then $xHx^{-1}$ also has index $p$. Therefore, $$xNx^{-1} = x\left(\bigcap_{[G:H]=p}H\right)x^{-1} = \bigcap_{[G:H]=p}xHx^{-1} = N.$$
To prove the coset of $x$ has exponent $n$, note that for every $H$ of index $p$, $G/H$ has order $p$, so $(xH)^p=eH$; therefore, $x^p\in H$.
Thus, $x^p$ lies in every subgroup of index $p$, hence in their intersection,