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Show that an abelian group of order 75 has a cyclic subgroup of order 15.

Do I need to use the fundamental theorem of finite abelian groups in some way?

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Yes, you can use the fundamental theorem of finite ablian groups. What do you get? –  Christoph Feb 26 at 16:15
"Can" is not same as "need to": no @Nicholas, you do not need to use that theorem...but you can, of course. –  DonAntonio Feb 26 at 17:31

2 Answers 2

By Cauchy's theorem, there is an element $u$ of order $3$ and an element $v$ of order $5$. Since $3$ and $5$ are coprime and the group is abelian, $uv$ has order $3\cdot 5=15$.

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Drats! So close. –  Pedro Tamaroff Feb 26 at 16:19

Yes, you can certainly use the Fundamental Theorem of finitely generated abelian groups to list all possible non-isomorphic abelian groups of order $75$.

To help distinguish between those abelian groups that are isomorphic vs. those that are not isomorphic, use the fact that $$\mathbb Z_{mn} \cong \mathbb Z_m\times \mathbb Z_n \iff \gcd(m, n) = 1$$

Then show that each of them necessarily has a subgroup of order $15$.

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The fundamental theorem is too much, really. –  Pedro Tamaroff Feb 26 at 16:19
This argument is over. Criticism should be welcomed, but personal attacks will not be condoned. –  Alexander Gruber Feb 26 at 17:47
By the way, I think this answer is fine- it's how I would do it. Even if it is overkill, working with abelian groups is very different from working with groups in the general sense. I think it's good that both perspectives are represented in the answers here. –  Alexander Gruber Feb 26 at 17:56
@AlexanderGruber Thanks. (It's the route that first popped in my head.) –  amWhy Feb 26 at 17:58

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