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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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  • $\begingroup$ Yes, you can use the fundamental theorem of finite ablian groups. What do you get? $\endgroup$ – Christoph Feb 26 '14 at 16:15
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    $\begingroup$ "Can" is not same as "need to": no @Nicholas, you do not need to use that theorem...but you can, of course. $\endgroup$ – DonAntonio Feb 26 '14 at 17:31
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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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    $\begingroup$ The fundamental theorem is too much, really. $\endgroup$ – Pedro Tamaroff Feb 26 '14 at 16:19
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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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    $\begingroup$ Drats! So close. $\endgroup$ – Pedro Tamaroff Feb 26 '14 at 16:19

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