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$G$ 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?

user1729
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Tumbleweed
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  • The order of the conjugacy class has to divide the order of the group by the class equation. From what you've said, they don't live in the centre. Therefore the conjugacy class has to be of size $3$ or $5$ (it can't be $15$ because the action of conjugation isn't transitive). Why can't the element of order $3$ live in a conjugacy class of order $5$? – Ian Coley Sep 18 '13 at 05:16
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    any group of order $15$ is abelian, see here – leshik Sep 18 '13 at 05:28
  • Lemma: If $|G|=pq$ in which $p<q$ are distinct primes then if $G$ is non-abelian then $$p|q-1$$ and moreover, $G$ has exactly $q$, $q-$ sylow subgroups. – Mikasa Sep 18 '13 at 06:08
  • I can prove that non-abelian groups of order 15 have conjugacy classes of order 7 as well. IOW, see the comments by leshik and Babak. – Jyrki Lahtonen Sep 18 '13 at 06:20
  • you have proved a group of order $15$ has trivial center??? :O HOW???? –  Sep 18 '13 at 07:10

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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.

Mark Bennet
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