# exponent of an abelian group

Let $p$ be a prime. Let $H_{i}, i=1,...,n$ be normal subgroups of a finite group $G$. I want to prove the following: If $G/H_{i}$, $i=1,...,n$ are abelian groups of exponent dividing $p−1$, then $G/N$ is abelian group of exponent dividing $p−1$ where $N=\bigcap H_{i},i=1,...,n$.

Proof: Since $G/H_{i}$, $i=1,...,n$ are abelian groups, then $G^{′}$ (the derived subgroup of $G$) is contained in every $H_{i}, i=1,...,n$. Hence $G^{′}$ is contained in $N$. Therefore $G/N$ is abelian. I do not know how to deal with the exponent.

Since all of the $G/H_i$ have exponent dividing $p-1$, we must have $g^{p-1}\in H_i$ for all $g\in G$. Therefore... – Miha Habič Aug 17 '12 at 4:00
@MihaHabič Therefore $g^{p-1} \in N$ for all $g \in G$. Thus $G/N$ has the exponent $p-1$ at most. am I right? – user28083 Aug 17 '12 at 4:07
That's right. $p$ being prime doesn't seem to matter. – Miha Habič Aug 17 '12 at 4:09
@MihaHabič Sorry, how did we know that the exponent of $G/N$ divides $p-1$? – user28083 Aug 17 '12 at 4:23
Somewhat more sophisticated but easy to show: (by "Poincare's Lemma") $G/N$ injects in the direct product $G/H_1 \times ... \times G/H_n$. The latter product is abelian and has exponent dividing $p-1$ and you are done.