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?

 

[blue_raver22]

I would like to commend you on your work so far and offer some additional insight into the proof you have provided.

Firstly, let's define some terms for clarity. An abelian group is a group where the group operation is commutative, meaning that the order in which elements are multiplied does not affect the result. The exponent of a group is the smallest positive integer n such that a^n = e, where a is any element of the group and e is the identity element. In other words, the exponent is the smallest number of times an element must be multiplied by itself to get the identity element.

Now, to prove that G/N is an abelian group of exponent dividing p-1, we need to show that for any element a in G/N, (a^(p-1)) = e. This will demonstrate that the exponent of G/N divides p-1.

As you have correctly stated, G^{'} is contained in N, meaning that any element in G^{'} can be written as a product of elements in N. Since G^{'} is the derived subgroup, it is generated by commutators of elements in G, which can be written as a product of elements in N. This means that every element in G^{'} can be written as a product of elements in N, and therefore, (a^(p-1)) can be written as a product of elements in N.

Now, we know that the exponent of each G/H_{i} divides p-1, which means that for any element b in G/H_{i}, (b^(p-1)) = e. Since G/H_{i} is a quotient group, every element in G can be written as a product of elements in G/H_{i}. This means that (a^(p-1)) can also be written as a product of elements in G/H_{i}. Since each G/H_{i} is abelian, the order of elements in the product does not matter, and we can rearrange the elements to show that (a^(p-1)) is equal to the identity element in G/H_{i}. Therefore, (a^(p-1)) = e.

Combining this with the fact that (a^(p-1)) can be written as a product of elements in N, we can conclude that (a^(p-1)) = e for any element a in G/N. This proves that the