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Let G be an abelian group of order 175 (=5*5*7). Assume $x^5=e$ has at least seven solutions. What is G isomorphic to?

I see and can show that G is isomorphic to the its Sylow subgroups (orders 7 and 25), and I know that finite abelian groups are direct products of cyclic groups. How can I combine these?

Can I say that the order 7 Sylow subgroup is isomorphic to a cyclic subgroup or order 7 (using that finite abelian groups are direct products of cyclic groups)?

And then is there a way to break the order 25 Sylow group into the product of 2 cyclic groups each of order 5 so that G is isomorphic to the product of an order 7 cyclic group and two order 5 subgroups?

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Yes, any (abelian or non abelian) group of order $7$ is cyclic, and this is true for any group of prime order. Now, an abelian group of order $25$ can be only $C_{25}$ or $C_5\times C_5$, so that your group $G$ can be $G=C_{25}\times C_{7}$ or $G=C_{5}\times C_{5}\times C_{7}$. Now you have to compute the number of elements of order $5$ in these two groups, and see which one is the right one.

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  • $\begingroup$ There are only 5 such elements in $C_{25}$ so I can rule that one out and so G is isomorphic to the product of an order 7 cyclic group and two order 5 subgroups. Thanks! $\endgroup$ – user143405 Apr 15 '14 at 20:53
  • $\begingroup$ You're welcome! $\endgroup$ – Daniele A Apr 15 '14 at 20:54

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