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It is given that the order of some finite abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

It is clear that since order of group is divisible by 10. By converse to Lagrange's Theorem, if 10 divides the order of the group G, then G has a subgroup of order 10.

But to ensure that this subgroup is cyclic.

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    @ArturoMagidin: Thanks$a$lot for your help.2011-10-10

3 Answers 3

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Hint:If $p$ and $q$ are distinct primes then $\mathbb{Z}_p\times\mathbb{Z}_q\cong \mathbb{Z}_{pq}$.

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    Okay, thanks. Unfortunately I have a few similar question to the asker but do not have the FTAG yet however.2014-08-07
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$G$ has a subgroup of order $2$ and order $5$ by Cauchy's theorem. Since $2$ is prime to $5$, the order of the product of two generators of these groups is $10$ and it generates then a (cyclic) subgroup of $G$ of order $10$.

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    Note that, to conclude that the product of an element of order 2 and an element of order 5 has order 10, it is very important that the group be abelian. Otherwise the order of the product could be anything.2011-10-10
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Hint: What abelian groups of order 10 can you think of?

(There is only one, up to isomorphism.)

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    @Tav: Precisely!2011-10-10