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).
Yes, you are on the right track, kind of! :)
If $G$ has an element of order $8$, then clearly $G$ is cyclic, so assume $G$ has no element of order $8$.
If $G$ also has no element of order $4$, then by Lagrange's theorem you have that every non-unit element of $G$ is of order $2$, and so $2G=0$. It then follows that $G$ is a $\mathbb{F}_2$ vector space, and thus, as a group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^3$.
So, we may assume that $g\in G$ has order $4$. Now, let $N=\langle g\rangle$. Choose $x\in G-N$ with $|x|=2$. Note that this is indeed possible. For, if not, then $G$ would have $6$ elements of order $4$, and $1$ element of order $2$. But, by a counting argument this is impossible. Let $K=\langle x\rangle$. Note then that $K\cap N=\{0\}$ (why?), $K,N\unlhd G$, and $KN=G$, since
$$\#(KN)=\frac{|K||N|}{|K\cap N|}=\frac{4\cdot 2}{1}=8$$
Thus, we have that
$$G\cong N\times K\cong (\mathbb{Z}/4\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$$
Best Answer
We just have to use the fundamental theorem for finite abelian groups.
Note $100=2^2\cdot 5^2$
You just have to choose a factorization of $2^2$ and a factorization of $5^2$.
The only two factorizations for $2^2$ are $2\cdot 2$ and $4$
the only two factorizations for $5^2$ are $5\cdot 5 $ and $25$.
So there are $4$ combinations, these give us all the groups:
$\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_5\times \mathbb Z_5$
$\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_{25}$
$\mathbb Z_4\times \mathbb Z_{5}\times \mathbb Z_5$
$\mathbb Z_4\times \mathbb Z_{25} \cong \mathbb Z_{100}$
The only of these that has maximum order $20$ is the third element in the list.