Classifying some abelian groups of order $2^5\times 3^5$ I'm requested to classify the abelian groups $A$ of order $2^5 \times 3^5 $ where :


*

*$| A/A^4 | = 2^4 $

*$ |A/A^3 | = 3^4 $
I need to write down the canonical form of each group . 
My question is, what does it mean  $| A/A^4 |$ ? 
I understand that the meaning is to:
$$G⁄H= \{Hg \mid  g\in G\}$$  
but the usage of it for classifying the groups is not clear to me. 
How does it help me with this case ? 
Regards
EDIT:
Suppose that I say $A=B \times C$ where $|B|=2^5$ and $|C|=3^5$ and 
$|A|=2^5 * 3^5$ . 
Now , the goals are :


*

*$|A|/|A|^3 = (2^5 * 3^5) /|A|^3 = 3^4 $ 


:meaning I need to make $|A^3|=3*2^5$


*

*$|A|/|A|^4 = (2^5 * 3^5) /|A|^4 = 2^4 $ 


:meaning I need to make $|A^4|=2*3^5$
So now after A=BC , how can I find the exact $|A^4|$ and $|A^3|$ ? 
thanks again 
 A: For any abelian group $A$, written multiplicatively, $A^n = \{a^n\mid a\in A\}$ is a subgroup (since $a^n(b^n)^{-1} = (ab^{-1})^n$ and $1^n=1\in A^n$). Therefore, we can talk about the quotient $A/A^n$, and we can talk about its size as a set. $|A/A^4|$ is the size of the group $A/A^4$, and $|A/A^3|$ is the size of the group $A/A^3$.
Let $A$ be an abelian group of order $2^5\times 3^5$. Then we know, from the Fundamental Theorem of Finitely Generated Abelian Groups, that we can express $A$ uniquely as
$$A \cong \mathbb{Z}_{2^{a_1}}\oplus\cdots \oplus\mathbb{Z}_{2^{a_k}} \oplus \mathbb{Z}_{3^{b_1}}\oplus\cdots\oplus\mathbb{Z}_{3^{b_{\ell}}},$$
where $1\leq a_1\leq\cdots\leq a_k$, $1\leq b_1\leq\cdots\leq b_{\ell}$, and $a_1+a_2+\cdots+a_k = 5$, $b_1+b_2+\cdots+b_{\ell} = 5$.
So for instance, one possibility might be
$$A \cong \mathbb{Z}_2 \oplus\mathbb{Z}_{2^4} \oplus \mathbb{Z}_{3^2}\oplus\mathbb{Z}_{3^3}.$$
But we want only some of the groups, namely, the ones where we also have $|A/A^4| = 2^4$ and $|A/A^3|=3^4$. Does this group satisfy this condition?
Well, since $(B\oplus C)^n = B^n\oplus C^n$, we can deal with each cyclic factor separately. Note that $(\mathbb{Z}_2)^4 = \{1\}$, and $(\mathbb{Z}_{2^4})^4 = \mathbb{Z}_{2^2}$; whereas, since $4$ is relatively prime to $3$, $(\mathbb{Z}_{3^2})^4 = \mathbb{Z}_{3^2}$, $(\mathbb{Z}_{3^3})^4 = \mathbb{Z}_{3^3}$.
So we have:
$$\begin{align*}
A &= \mathbb{Z}_2 \oplus \mathbb{Z}_{2^4} \oplus \mathbb{Z}_{3^2}\oplus\mathbb{Z}_{3^3}\\
A^4 &= (\mathbb{Z}_2)^4 \oplus (\mathbb{Z}_{2^4})^4 \oplus (\mathbb{Z}_{3^2})^4 \oplus (\mathbb{Z}_{3^3})^4\\
&\cong \{1\} \oplus \mathbb{Z}_{2^2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{3^3}.
\end{align*}$$
Therefore, since we are dealing with finite groups,
$$\left|\frac{A}{A^4}\right| = \frac{|A|}{|A^4|} = \frac{2^5\times 3^5}{2^2\times 3^5} = 2^3,$$
so this $A$ does not satisfy the condition. 
You are asked to find all the ones that do.
Hint. Prove that:


*

*If $p$ and $q$ are primes, $p\neq q$, then $(\mathbb{Z}_{p^a})^{q^b} = \mathbb{Z}_{p^a}$;

*If $p$ is a prime, and $a\leq b$, then $(\mathbb{Z}_{p^a})^{p^b} = \{1\}$.

*If $p$ is a prime, and $a\gt b$, then $(\mathbb{Z}_{p^a})^{p^b}\cong \mathbb{Z}_{p^{a-b}}$.

