The first part of the problem asks you to prove that an abelian group $G$ with order $100$ must contain an element of order $10$. For this part, I use Sylow theorem to list possiblities for $H$ and $K$ where $|H|$=$2^{2}$ and $|K|$=$5^{2}$. $K$ must be normal since there is not subgroup of order $25$ while $H$ might not be normal since $1$\equiv$25$ mod $2$. Also I proved that $H$$K$=$G$ and they have only the identity element in common.
Then, if $H$ and $K$ are normal. G must be isomorphic to one of the following groups:
$Z_{4}$ $\times$ $Z_{25}$
$Z_{2}$ $\times$ $Z_{2}$ $\times$ $Z_{25}$
$Z_{4}$ $\times$ $Z_{5}$ $\times$ $Z_{5}$
$Z_{2}$ $\times$ $Z_{2}$ $\times$ $Z_{5}$ $\times$ $Z_{5}$
From above, it is easily to pick up element of order $10$ for each. But my confusion is that since $G$ is an abelian group, how can I use the theorem that any finite abelian group is isomorphic to a direct product of cyclic groups?
Also, since the second part asks you if no element of $G$ has order greater than $10$, what are its torsion coefficients? I think my way of listing the possibilities are too complicated. Are there any more explicit ways to solve the problem?
Thanks a lot.