# 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

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No, your goals are $|C|/|C^3| = 3^4$ and $|B|/|B^4| = 2^4$. Now, write $B$ as a direct sum of cyclic $2$-groups (that is, cyclic groups of order a power of $2$), and see what $B^4$ is; write $C$ as a direct sum of cyclic groups of order a power of $3$, and see what $C^3$ is. You need $|B^4| = 2$, and you need $|C^3| = 3$ – Arturo Magidin Mar 18 '12 at 3:45
@ArturoMagidin: But how did you get to $|C/C^3|$ and $|B/B^4|$ ? What I mean is , how do you know what is $|A^3|$ and $|A^4|$ ? – ron Mar 18 '12 at 5:15
As was already explained, $A = B\times C$, so $A^k = B^k\times C^k$ for any $k$. Since the order of $C$ is relatively prime to $2$, then $C^4 = C$, so $A/A^4 = (B/B^4)\times (C/C^4) = (B/B^4)\times (C/C) \cong B/B^4$. And since the order of $B$ is relatively prime to $3$, then $B^3=B$, so again we have $A/A^3 = (B/B^3)\times (C/C^3) \cong C/C^3$. If there was no simplification in considering $B$ and $C$, then what's the point of introducing them in the first place? If you didn't understand why you should introduce them, then you should have asked, not just put them in and then ignore them. – Arturo Magidin Mar 18 '12 at 5:21
I lost you with $C^4 = C$ , can you please explain ? – ron Mar 18 '12 at 5:23
Look at the hint I wrote in my answer; the one you've been studiously not trying to prove. But even more generally: if the order of $g$ is $n$, and $k$ is relatively prime to $n$, then $g$ is a $k$th power: write $1 = an+bk$ for some integers $a$ and $b$. Then $g = g^1 = g^{an+bk} = (g^n)^a(g^b)^k = 1^a(g^b)^k = (g^b)^k$. Since every element of $C$ has order relatively prime to $4$, every element of $C$ is the fourth power of someone in $C$. That means that $C\subseteq C^4$; since $C^4\subseteq C$ always holds, you get equality. – Arturo Magidin Mar 18 '12 at 5:26

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 $(\mathbb{Z}_{p^a})^{p^b}\cong \mathbb{Z}_{p^{a-b}}$.
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I need to have some thinking regarding this question , it requires a lot of theory that I don't fully have at the moment . I hope I'd some answers in the next few hours . thanks ! – ron Mar 17 '12 at 6:03
@ron: If you can establish the properties I listed in the HINT, then it turns out to be a fairly straighforward proposition. But you really need to get your hands "dirty" on this one and play around with a few examples of groups to see how things play out. – Arturo Magidin Mar 17 '12 at 20:37
I have some calculations that I've made . Where can I write them ? here in the comments , or re-edit my original post? – ron Mar 18 '12 at 1:38
@ron: Comments are difficult; adding them to the original post (without modifying the question) would work. – Arturo Magidin Mar 18 '12 at 2:19
I've added the changes , not much but I think it would do . But I'm still missing something there... – ron Mar 18 '12 at 3:44