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).
There's not a simple formula known for this, and certain aspects of the question are the subject of ongoing research. For example, this paper summarizes some research that has been done on the maximum possible order for an element of $S_n$.
In general, the way to find the number of elements of order $k$ in $S_n$ is:
Determine all possible cycle types for an element of order $k$, and then
Determine the number of elements having each of these cycle types.
For step (1), you're just looking for all possible ways to partition $n$ into cycles so that the least common multiple of the cycle lengths is $k$. For example, if permutation has order six, then all the cycles must have length $1$, $2$, $3$, or $6$, with either at least one $6$-cycle or one cycle each of lengths $2$ and $3$. So if we want to count the number of permutations of order six in $S_8$, the possibilities are
- One $6$-cycle, one $2$-cycle,
- One $6$-cycle, two $1$-cycles,
- Two $3$-cycles, one $2$-cycle,
- One $3$-cycle, two $2$-cycles, and one $1$-cycle, or
- One $3$-cycle, one $2$-cycle, and three $1$-cycles.
Step (2) is easy once you figure out step (1). In particular, the number of permutations in $S_n$ with a given cycle structure is
$$
\frac{n!}{\prod_{d=1}^n (c_d)!\,d^{c_d}}
$$
where $c_d$ denotes the number of cycles of length $d$. For example, the number of elements of $S_{20}$ having four $1$-cycles, five $2$-cycles, and two $3$-cycles is
$$
\frac{20!}{(4!\cdot 1^4)(5!\cdot 2^5)(2!\cdot 3^2)} \;=\; 1\text{,}466\text{,}593\text{,}128\text{,}000.
$$
The following table shows the number of elements of each order in $S_2$ through $S_8$.
$$
\begin{array}{crrrrrrrrrrr}
& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 10 & 12 & 15 \\
\hline
S_2 & 1 & 1 & - & - & - & - & - & - & - & - & -\\
S_3 & 1 & 3 & 2 & - & - & - & - & - & - & - & - \\
S_4 & 1 & 9 & 8 & 6 & - & - & - & - & - & - & - \\
S_5 & 1 & 25 & 20 & 30 & 24 & 20 & - & - & - & - & - \\
S_6 & 1 & 75 & 80 & 180 & 144 & 240 & - & - & - & - & - \\
S_7 & 1 & 231 & 350 & 840 & 504 & 1470 & 720 & - & 504 & 420 & - \\
S_8 & 1 & 763 & 1232 & 5460 & 1344 & 10640 & 5760 & 5040 & 4032 & 3360 & 2688
\end{array}
$$
This table is entry A057731 at OEIS.
Best Answer
By chinese remainder theorem $\mathbb Z_{100}^*\cong \mathbb Z_4^*\times \mathbb Z_{25}^*$.
An element of the latter subgroup $(a,b)$ has order $lcm(|a|,|b|)$ where $|a|$ and $|b|$ are the orders in $\mathbb Z_4^*$ and $\mathbb Z_{25}^*$. So if $(|a|,|b|)=20$ we must have $|b|=20$.
Of course it is a well known theorem $\mathbb Z_{p^\alpha}^*$ is cyclic for $p$ an odd prime.
So the pair $(a,b)$ has order $20$ if and only if $|b|=20$. Since $\mathbb Z_{25}^*\cong \mathbb Z_{20}$ there are $\varphi(20)=8$ suitable options for $b$ (cyclic groups have $\varphi(n)$ elements of order $n$)and hence $2$ options for $a$ ($\mathbb Z_{4}^*$ has two elements).
Hence there are $16$ elements of order $20$.
If you want to find them all then one way to do it is by finding a generator for $\mathbb Z_{25}^*$
Lets try with $2$, we get $2,4,8,16,7,14,3,6,12,24,23,21,17,9,18,11,22,19,13,1$
So in fact $2$ generates. From here we can see the groups as:
$2,2^2,2^3,2^4,2^5\dots 2^{20}$, of course we want the elements in which the exponent is relatively prime with $20$. So we want $2^{1},2^3,2^7,2^9,2^{11},2^{13},2^{17},2^{19}$ which corresponds to $2,8,3,12,17,22,13$.
Each of these values for $b$ shall give us two congruence classes $\bmod 100$ of order $20$, they are $(1,b)$ and $(3,b)$ respectively. It is clear these numbers are $b,b+50$ if $b$ is odd and $b+25,b+75$ if $b$ is even. Doing this for each $b$ is all that remains, lets do it:
$27,77$ results from $b=2$ and $33,83$ result from $b=8$. The other $12$ are:
$3 ,53, 37, 87, 23, 73, 17, 67, 47, 97, 13, 63$. So these are the $16$ elements.