# Abelian groups and exponent of a 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.

Related Linear and Abstract Algebra News on Phys.org
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.

I think you only need the following two things:

1) If A is an abelian group of exponent m, then the exponent of any subgroup of A and of any homomorphic image of A divides m

$$(2)\,\,\forall i\,\,,\,\, N \leq H_i\, \Longrightarrow \,\,\forall\,\, g\in G\,\,,\, m(g + N) = N\Longrightarrow mg \in N \leq H_i\Longrightarrow \,m(g + H_i) = H_i \Longrightarrow$$
$$\Longrightarrow\, m\,\, \text{divides the exponent of}\,\, G/H_i\,\, .$$

DonAntonio

thank you very much.

Excuse me, how did you know that m is less than the exponent of G/Hi?