Abelian groups and exponent of a group

[moont14263]

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.

 

[DonAntonio]

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.

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

 

[moont14263]

thank you very much.

 

[moont14263]

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