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).
See the Wikipedia article on finitely-generated abelian groups for more information. We have
Theorem A. Any finite abelian group $G$ admits a unique invariant factor decomposition.
Theorem B. Any finite abelian group $G$ admits a unique primary decomposition.
These decompositions are different, but logically Thm A $\Leftrightarrow$ Thm B: that is, once you prove one, the other follows from Sun-Ze (aka CRT), ${\bf Z}/nm\cong {\bf Z}/n\times{\bf Z}/m$ when $(n,m)=1$. Further one can compute one decomposition from the other (+ vice-versa). Let's discuss how.
Suppose we have the invariant factor decomposition of $G$ from Thm A as follows:
$$G\cong \frac{\bf Z}{n_1}\times\frac{\bf Z}{n_2}\times\cdots\times\frac{\bf Z}{n_l}.$$
Then each $n_i$ factors as $\prod_pp^{e(i,p)}$ for primes $p$ and exponents $e(i,p)$. By SZ then
$$G\cong\prod_{i=1}^l\frac{\bf Z}{n_l}\cong\prod_{i=1}^l\prod_p\frac{\bf Z}{p^{e(i,p)}}\cong\prod_p\left[\prod_{i=1}^l\frac{\bf Z}{p^{e(i,p)}}\right].$$
Notice the factors $\prod_{i=1}^l{\bf Z}/p^{e(i,p)}$ are $p$-groups; this is the primary decomposition, which is the same as the Sylow decomposition (as a direct product of Sylow $p$-subgroups, available for all nilpotent groups) when our group is abelian. It is probably best to think of Sylow theory as the inevitable noncommutative generalization of arithmetic in the theory of finitely-generated abelian groups.
Our original hypothesis that we had the invariant factor decomposition means that $n_i\mid n_{i+1}$, so we must have $e(i,p)\le e(i+1,p)$ for each $p$. Also, some of the $e$'s may be $0$. Thus if we begin with a primary decomposition as regarded in Thm B, we may order the prime exponents in an increasing order, and fill in extra $0$s as needed so that each prime has the same number of exponents, then form the $n_i$ out of the prime powers created from these exponents.
Here's an example. From a primary decomposition we obtain
$$\begin{array}{llccccc} G & \cong & \left(\frac{\bf Z}{2} \times \frac{\bf Z}{2}\times\frac{\bf Z}{8}\right) & \times & \left(\frac{\bf Z}{3}\times\frac{\bf Z}{27}\right) & \times & \left(\frac{\bf Z}{5}\right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}} \times \color{Blue}{\frac{\bf Z}{2}} \times \color{Green}{\frac{\bf Z}{8}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{3}} \times \color{Green}{\frac{\bf Z}{27}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{1}} \times \color{Green}{\frac{\bf Z}{5}} \right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}\times\frac{\bf Z}{1}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Blue}{\frac{\bf Z}{2}\times\frac{\bf Z}{3}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Green}{\frac{\bf Z}{8}\times\frac{\bf Z}{27}\times\frac{\bf Z}{5}} \right) \\ & \cong & \frac{\bf Z}{2\times1\times1} & \times & \frac{\bf Z}{2\times3\times1} & \times & \frac{\bf Z}{8\times27\times5} \\[7pt] & \cong & {\bf Z}/2 & \times & {\bf Z}/6 & \times & {\bf Z}/1080 \end{array}$$
as our invariant factor decomposition, with $2\mid6\mid1080$. And to see how to go in reverse with this same example, just read from bottom to top.
In conclusion, then, there are two standard fundamental representations of a finite abelian group, the invariant-factor decomposition and the primary decomposition. Theorems A and B state their existence and uniqueness; these are roughly logically equivalent using the remainder theorem, and it is relatively easy to move between the two decompositions using prime factorization. The IF representation invokes linear algebra, the primary decomposition invokes group theory (it is positioned within Sylow theory), and they are both positioned within commutative algebra.
Best Answer
We are using Cauchy's Theorem for the existence of an element $h$ of order $11$ and an element $k$ of order $13$. Let $H=\langle h \rangle$ and $K=\langle k \rangle$. Note that $H \cap K=1$ (use Lagrange's Theorem).
Step 1: $K \lhd G$.
Proof If $K$ is not normal we can find a $g \in G$, with $K^g:=g^{-1}Kg \neq K$. Note that, as $K$, also $K^g$ is a subgroup of order $13$ and by Lagrange $K \cap K^g=1$ (here we use that $K^g \neq K$). Hence $$143=|G| \geq |KK^g|=\frac{|K|\cdot|K^g|}{|K \cap K^g|}=|K|\cdot |K^g|=13 \cdot 13 =169$$ a contradiction. $\square$
Step 2: $K \subseteq Z(G)$.
Proof Since $K$ is normal, $G$ acts by conjugation on $K$ leading to a homomorphic embedding $G/C_G(K) \hookrightarrow \text{Aut}(K)$. But $\text{Aut}(K) \cong C_{12}$. So $|G/C_G(K)|$ divides both $12$ and $143$, hence $G=C_G(K)$, equivalent to $K$ being central in $G$. $\square$
Step 3: $G=HK$ (and $H \lhd G$)
Proof $|HK|=\frac{|H|\cdot |K|}{|H \cap K| }= 11 \cdot 13=143$. It follows that $G=HK$. Of course $H$ normalizes $H$ and $K \subseteq Z(G)$ normalizes $H$, whence $H$ is normal in $G$. $\square$
Basically you are done now, since every pair of elements of $G$ commute, indeed $H$, $K$ are abelian and $K$ is central. But one can go one step further.
Step 4: $G \cong H \times K \cong C_{11} \times C_{13} \cong C_{143}.$
Proof (sketch) Every $g \in G$ can be uniquely written as $g=hk$ with $h \in H, k \in K$. This amounts to a bijective homomorphism $G \rightarrow H \times K$, by sending $g$ to $(h,k)$. $\square$
Remark It is well-known that if $n$ is a natural number, then there is only one group of order $n$ if and only if $\text{gcd}(\varphi(n),n)=1$. Of course this group needs to be isomorphic to $C_n$.