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.
Best Answer
$G$ 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 $10$.