An abelian group of order 100

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.

Solution 1:

If you are assuming that $G$ is an abelian group of order $100$, then you don't need Sylow's Theorems, you just need Cauchy's Theorem: since $2$ and $5$ divide $|G|$, $G$ has an element $a$ of order $2$, and an element $b$ of order $5$. Then $\langle a\rangle\cap\langle b\rangle = \{0\}$ (writing the groups additively), so $a+b$ has order $10$, as is easily verified: $$\begin{align*} k(a+b) = 0 &\iff ka+kb = 0\\ &\iff ka=-kb\\ &\iff ka,kb\in\{0\}\\ &\iff 2|k\text{ and }5|k\\ &\iff 10|k. \end{align*}$$

The second part asks you to whether there are no elements of order greater than 100 in an abelian group of order 100? The cyclic group of order 100 shows that this need not be the case. In fact, the only abelian group of order $100$ in which there are no elements of order greater than $10$ is the group $\mathbf{Z}_2\oplus\mathbf{Z}_2\oplus\mathbf{Z}_{5}\oplus\mathbf{Z}_{5}\cong \mathbf{Z}_{10}\oplus\mathbf{Z}_{10}$.

Or I may be misunderstanding the seecond part, and instead you are told that $G$ has no elements of order greater than $10$... the only possible orders, by Lagrange's Theorem, are $1$, $2$, $4$, $5$, $10$, $20$, $25$, $50$, and $100$. Since you are told there are no elements of order greater than $10$, then the orders must be $1$, $2$, $4$, $5$, or $10$. But if you have an element $x$ of order $4$, then $x+b$ is of order $20$ (same argument as above), a contradiction. So every element is of order $1$, $2$, $5$, or $10$. And there are certainly elements of order $1$ (namely, $0$), order $2$ and $5$ (Cauchy's Theorem), and order $10$ (first part of the problem).