Originally Posted by

**anlys** Hello Tonio,

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

Yes

For instance, can G be written as $\displaystyle G=C_8\times C_9\times C_ (25)$ as well? So, in this case, isn't its maximal cyclic subgroup is $\displaystyle 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 $\displaystyle p^3q^2r^2$ what's the size of the maximal cyclic subgroup in GENERAL. Of course, if we happen to meet the cyclic group of order $\displaystyle p^3q^2r^2$ 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 $\displaystyle p^3q^2r^2$ 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 $\displaystyle p^3q^2r^2$, 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 $\displaystyle p^2rq$? Isn't it greater than prq?

Again, in MY example, the sbgp. of that order is $\displaystyle C_p\times C_p\times C_r\times C_q$, 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.