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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$2012-03-18
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    @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 $|A^4|$ ?2012-03-18
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    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.2012-03-18
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    I lost you with $C^4 = C$ , can you please explain ?2012-03-18
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    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.2012-03-18
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    @ArturoMagidin : Something is not clear to me , can I say that $|A^4|=2*3^5$ ?2012-03-18
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    $|A^4| = |(B\times C)^4| = |B^4\times C^4|= |B^4|\times|C^4| = |B^4|\times|C| = |B^4|\times 3^5$. Whether you can say that it equals $2\times 3^5$ depends on what $B$ is.2012-03-18

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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 !2012-03-17
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    @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.2012-03-17
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    I have some calculations that I've made . Where can I write them ? here in the comments , or re-edit my original post?2012-03-18
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    @ron: Comments are difficult; adding them to the original post (without modifying the question) would work.2012-03-18
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    I've added the changes , not much but I think it would do . But I'm still missing something there...2012-03-18