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).
A very simple example: let G be the group of permutations of $\mathbf Z$. Denote by $s$ the ‘symmetry’ $x\mapsto -x$ and $t$ be the ‘translation’ $x\mapsto x+1$. $s$ is of order 2, but $t$ has infinite order – indeed, $t^k$ is simply $x\mapsto x+k$. Now, it's easy to check $s\circ t$ has order 2 like $s$, but $s\circ(s\circ t)=t$ has infinite order.
Best Answer
The original Heisenberg group is the group you describe: the $3\times 3$ matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right)\tag{1}$$ with $a,b,c\in\mathbb{R}$, which form a group under the usual matrix multiplication.
However, there is a generalization, and they are all generally known as "the Heisenberg group", and one is supposed to know which by context. Given any field $F$ (for example, the rationals, the complex numbers, the reals, the algebraic numbers, or a finite field of order $p$), you consider the matrices of the form $(1)$, but where now $0$, $1$, and $a,b,c$ are taken in the field $F$.
You should verify that these matrices form a group. In particular, if you take the entries in $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}$, the integers modulo $3$, you obtain a nonabelian group of order $27$. (In general, taking them in $\mathbb{Z}/p\mathbb{Z}$ gives you a nonabelian group of order $p^3$). You should verify that it has the property you want.