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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.

Thanks in advance.

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    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...2012-08-17
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    @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?2012-08-17
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    That's right. $p$ being prime doesn't seem to matter.2012-08-17
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    @MihaHabič Sorry, how did we know that the exponent of $G/N$ divides $p-1$?2012-08-17
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    The exponent divides all other powers which kill every element. This is basically the definition and you should prove it if you don't know it.2012-08-17
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    what dose mean exponent of group G???2013-09-04

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