How to show that the group is abelian? I have this exercise:

a. Let $\sigma \in S_{15}$ be an element of order 5. What type of
  cycles can occur in the decomposition of $\sigma$ in disjoint cycles?
b. Let $S \subseteq S_{15}$ be one of the Sylow 5-group in $S_{15}$.
  What is the order of S? Show that S is abelian and that all nontrivial
  elements are of order 5.
c. Let p be a prime number and let $0<k<p$ be a natural number. Show
  that the Sylow p-subgroups of the symmetric group $S_{kp}$ is abelian.

my attempt and questions:
a.In this part I think we only can get 5 cycles, because the order of a product of disjoint cycles is the least common multiple of the orders of all, and 5 is a prime number.
b.
Since the order of $S_{15}$ is 15!. We must have that the order of the Sylow 5-subgroup is $5^3=125$, since 5 occurs 3 times in the product of the order. But I am not sure how to show that S is abelian and that all nontrivial elements are of order 5. For all I know there might be elements of order 25 and 125. 15! is so big that there should be no problem for this to fit, so there must be another argument for this? And how do I show that S is abelian?
c. Here I have no idea. But I guess it is a generalisation of b. So if I knew b, maybe it would be easier.
 A: Your solution for part a is correct. (Or, at least, it is true that every cycle in the product of disjoint cycles must be a 5-cycle. There can, of course, be more than one cycle in the product.)
Now we look at the rest of the problem. I will prove part c because part b is a special case of it.
We want to show that if $p$ is a prime number, and $0<k<p$ is a natural number, then the Sylow $p$-subgroups of the symmetric group $S_{kp}$ are abelian.
First, note that the multiples of $p$ which are less than or equal to $kp$ are $p, 2p, 3p, ..., kp$, and each of these is divisible by $p$ exactly once since $k<p$. Thus the largest power of $p$ which divides $(kp)!$ is $p^k$, and so the Sylow $p$-subgroups of $S_{kp}$ have order $p^k$.
Let $G$ be a Sylow $p$-subgroup of $S_{kp}$. We will show that every element of $G$ has order $p$. To see this, let $x$ be an element of $G$, and consider the disjoint cycle structure of $x$. We note that the order of $x$ must be a divisor or $p^k$ and hence is a power of $p$. Thus the length of each cycle in the disjoint cycle structure of $x$ must be a power of $p$. But since $x$ is an element of $S_{kp}$, the length of each cycle can be at most $kp$, and hence is not divisible by $p^2$. (Since $k < p$). Thus the length of each cycle in the disjoint cycle structure of $x$ is $p$, and we see that $x$ has order $p$. (This wasn't actually required in part c, but was asked in part b, and so I have included it)
To show that $G$ is abelian, we will exhibit a Sylow $p$-subgroup of $S_{kp}$ which is abelian. The result then follows because all of the Sylow $p$-subgroups of $S_{kp}$ are isomorphic.
Consider the subgroup of $S_{kp}$ which is generated by the elements $x_1, x_2, ..., x_k$ where $x_m$ is the cycle $(((m-1)p+1)\,\, ((m-1)p+2)\,\,...\,\,mp)$. (e.g. $x_1$ is the cycle $(1\,\,2\,\,3\,\,...\,\,p)$, $x_2$ is the cycle $((p+1)\,\,(p+2)\,\,...\,\,2p)$, and so on)
We note that these cycles are disjoint, and hence commute.
We can see that the elements of this subgroup are precisely the elements
$$\prod_{i=1}^{k} x_i^{m_i}$$ 
where $0 \leq m_i < p$ for each $i$. We see that the group has $p^k$ elements, and hence is a Sylow $p$-subgroup of $S_{kp}$.
We can also see that the group is abelian, by noting that if
$$a = \prod_{i=1}^{k} x_i^{m_i}\quad\text{and}\quad b = \prod_{i=1}^{k} x_i^{n_i}$$
then
$$ab = \prod_{i=1}^{k} x_i^{m_i+n_i} = ba$$
where the last equality follows since the $x_i$'s commute.
