Nonabelian group of order 15 have conjugacy classes of order 5 and order 3? $G$ is a nonabelian group of order 15, and I already proved it has a trivial centre.
By Cauchy’s Theorem, $G$ must contain an element of order 3 and one of order 5 - but why these exist in conjugacy classes of order 5 and order 3?
 A: I'll sketch out some general facts about conjugacy classes which are useful to know, and which show why the sizes of the classes divide the order of the group.
Suppose $x\in G$. Consider the centraliser of $x$ in $G$. This starts as the set $C_x$ of elements $a\in G$ for which we have $ax=xa$.
Then notice that $1\in C_x$, the product of two elements of $C_x$ is also an element because in this case $abx=axb=xab$. Also if $ax=xa$ we have $xa^{-1}=ax^{-1}$, so $C_x$ is a subgroup of G.
Now let's think about when two conjugates of $x$ are equal. This happens precisely when $a^{-1}xa=b^{-1}xb$ for some $a,b \in G$, or when $ba^{-1}x=xab^{-1}$, or when $ba^{-1}\in C_x$, or when $b\in C_xa$ (check these are different ways of saying the same thing).
This argument is reversible, so we can say that the conjugates of $x$, which form a conjugacy class, are in one to one correspondence with the cosets of $C_x$. The number of conjugates, hence the size of the conjugacy class, is therefore $\frac {|G|}{|C_x|}$ and divides the order of the group.
A second fact which comes in surprisingly handy in a number of arguments is that the identity element always forms a conjugacy class of size $1$.
The element $x$ is in a conjugacy class of size $1$ if $C_x=G$ - this means that $x$ commutes with every element of $G$ ie $x$ is in the centre of $G$. In an abelian group all conjugacy classes have size $1$, and if all conjugacy classes have size $1$ the group is abelian.
