# Finite Abelian Groups

• November 8th 2009, 08:29 AM
anlys
Finite Abelian Groups
Let G be a finite Abelian group of order $p^3q^2r^2$ where p, q, and r are distinct primes. Define "size" as the number of prime factors (For example, $p^2q$ and $p^2r$ have size 3, but $q^2$ have the size 2, even though $q^2$ might be larger than both $p^2q$ and $p^2r$)

Question: What is the minimum size for "the" largest cyclic subgroup of G and which primes divide the size of this subgroup? Explain your answer.

I am not too sure how to do this problem. By the Fundamental Theorem of Finite Abelian Groups, there are 12 possible choices of groups that G is isomorphic to, right? I am really confused on how to find "the" largest cyclic group of G here. Any help would be greatly appreciated. Thank you in advance.
• November 8th 2009, 09:52 AM
tonio
Quote:

Originally Posted by anlys
Let G be a finite Abelian group of order $p^3q^2r^2$ where p, q, and r are distinct primes. Define "size" as the number of prime factors (For example, $p^2q$ and $p^2r$ have size 3, but $q^2$ have the size 2, even though $q^2$ might be larger than both $p^2q$ and $p^2r$)

Question: What is the minimum size for "the" largest cyclic subgroup of G and which primes divide the size of this subgroup? Explain your answer.

I am not too sure how to do this problem. By the Fundamental Theorem of Finite Abelian Groups, there are 12 possible choices of groups that G is isomorphic to, right? I am really confused on how to find "the" largest cyclic group of G here. Any help would be greatly appreciated. Thank you in advance.

Well, G has (at least one) a subgroup of each order dividing the order of G, so it has a sbgp of order prq which must be cyclic and of size 3, and it's divided by all the three primes dividing the order of G.
But for this I don't think we can say in general that any other group of order larger than pqr in G is cyclic:

for example, $G=C_2\times C_2\times C_2\times C_3\times C_3\times C_5\times C_5$ is a group of order $2^33^25^2$ and its maximal cyclic sbgp. is $C_2\times C_3\times C_5$

Tonio
• November 8th 2009, 10:44 AM
anlys
Quote:

Originally Posted by tonio
Well, G has (at least one) a subgroup of each order dividing the order of G, so it has a sbgp of order prq which must be cyclic and of size 3, and it's divided by all the three primes dividing the order of G.
But for this I don't think we can say in general that any other group of order larger than pqr in G is cyclic:

for example, $G=C_2\times C_2\times C_2\times C_3\times C_3\times C_5\times C_5$ is a group of order $2^33^25^2$ and its maximal cyclic sbgp. is $C_2\times C_3\times C_5$

Tonio

Hello Tonio,
Thank you for your response. In your example, for the group G of order $2^33^25^2$, isn't there are twelve possible ways we could express the direct product? For instance, can G be written as $G=C_8\times C_9\times C_ (25)$ as well? So, in this case, isn't its maximal cyclic subgroup is $C_ (1800)$? How do we know which of these twelve choices will give us the maximal cyclic subgroup? Also, you mentioned that for my problem, it has a sbgp of order prq which must be cyclic and of size 3, and it's divided by all the three primes dividing the order of G. But how do we know that subgroup of order prq is the largest subgroup? How about subgroup with order $p^2rq$? Isn't it greater than prq? Finally, how do we know all the three primes divided 3? Since 3 is a prime number, its divisors must be 1 and itslef only, right? Appreciate your clarification on this. Thank you, Tonio.
• November 8th 2009, 03:12 PM
tonio
Quote:

Originally Posted by anlys
Hello Tonio,
Thank you for your response. In your example, for the group G of order $2^33^25^2$, isn't there are twelve possible ways we could express the direct product?

Yes

For instance, can G be written as $G=C_8\times C_9\times C_ (25)$ as well? So, in this case, isn't its maximal cyclic subgroup is $C_ (1800)$?

No, this is not what you asked. There is not "its maximal cyclic..." since we're not fixing a group (otherwise the problem is trivial as we know EXACTLY what group we're talking about and we can check each case separately).
You asked for a general abelian group of order $p^3q^2r^2$ what's the size of the maximal cyclic subgroup in GENERAL. Of course, if we happen to meet the cyclic group of order $p^3q^2r^2$ then we're done, but that is NOT the general case.

How do we know which of these twelve choices will give us the maximal cyclic subgroup?

If this was the question the answer is trivial: the cyclic group of order $p^3q^2r^2$ is obviously the abelian group of that order with a cyclic sbgp. (it itself) of largest order

Also, you mentioned that for my problem, it has a sbgp of order prq which must be cyclic and of size 3, and it's divided by all the three primes dividing the order of G. But how do we know that subgroup of order prq is the largest subgroup?

In my example, one of the 12 abelian groups of order $p^3q^2r^2$, the one I gave is clearly the maximal cyclic sbgp. of the group, since ANY other sbgp. of that group will have a direct factor of order 2,3 or 5, making already NOT cyclic.

How about subgroup with order $p^2rq$? Isn't it greater than prq?

Again, in MY example, the sbgp. of that order is $C_p\times C_p\times C_r\times C_q$, which is NOT cyclic

Finally, how do we know all the three primes divided 3?

Isn't it clear that a group of order pqr is divisible by p, q and r??
If this question was taken from a book I think it'd be a good idea to check it again and, perhaps, write down the question EXACTLY as it shows there

Tonio

Since 3 is a prime number, its divisors must be 1 and itslef only, right? Appreciate your clarification on this. Thank you, Tonio.

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