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## Main Question or Discussion Point

Let [itex]p[/itex] be a prime. Let [itex]H_{i}[/itex], [itex]i=1,...,n[/itex] be normal subgroups of a finite group [itex]G[/itex]. I want to prove the following:

If [itex]G/H_{i}[/itex], [itex]i=1,...,n[/itex] are abelian groups of exponent dividing [itex]p-1[/itex], then [itex]G/N[/itex] is abelian group of exponent dividing [itex]p-1[/itex] where [itex]N=\bigcap H_{i} ,i=1,...,n[/itex].

Proof:

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

Thanks in advance.

If [itex]G/H_{i}[/itex], [itex]i=1,...,n[/itex] are abelian groups of exponent dividing [itex]p-1[/itex], then [itex]G/N[/itex] is abelian group of exponent dividing [itex]p-1[/itex] where [itex]N=\bigcap H_{i} ,i=1,...,n[/itex].

Proof:

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

Thanks in advance.