Conjugacy Classes of a group G - Intuitive Understanding How can I intuitively understand conjugacy classes of a group G. 
I feel I have a strong understanding of Equivalence Relations, and just completed the proof showing that conjugacy is an equivalence relation; however, when asked to compute the conjugacy classes of S3, I find myself with no intuition for where to start.
Best
 A: In general, given a finite group $G$, you can find its conjugacy classes by choosing an element $x$, and computing $g^{-1}xg$ for all $g$ in $G$; this will give you the conjugacy class of $x$. Then pick an element $y$ which isn't in the class of $x$, and compute $g^{-1}yg$ for all $g$ in $G$; this will give you the conjugacy class of $y$. If $G$ has any elements in neither of the classes you have now computed, then pick one of them, compute its conjugacy class as above, and keep doing this until you have exhausted the group. 
There are often shortcuts. David Ullrich has hinted that if your group is $S_n$ for some $n$, then two elements are in the same conjugacy class precisely when they have the same cycle structure. If $G$ is abelian, then each conjugacy class consists of a single element. The size of each conjugacy class is a factor of the size of the group. The elements of a conjugacy class all have the same order. If you know how big the centralizer of an element $x$ is, then you know how big its conjugacy class is, since the product of the two numbers is the order of the group. 
Take some small (nonabelian) groups, like $D_8$ and $A_4$ and $D_{10}$, and work out the conjugacy classes. You'll be glad you did. 
A: Major hint: Say $\phi_1,\phi_2\in S_5$ are conjugate by $\psi$ (that is, $\phi_2=\psi\phi_1\psi^{-1}$). Say the representation of $\phi_1$ as a product of disjoint cycles is $$\phi_1=(a,b,c)(d,e).$$Show that
$$\phi_2=(\psi(a),\psi(b),\psi(c))(\psi(d),\psi(e)).$$
Looks like one conjugacy class in $S_n$ for each way of writing $n$ as a sum of positive integers: $3=2+1=1+1+1$ means that $S_3$ has three conjugacy classes.
A: If $G$ is the group of transformations of an object $P$ (like a polytope) and the transformation $x$ is determined by some element of $P$ -- say it's the symmetry through the hyperplane $H$, then $yxy^{-1}$ is the symmetry through $yH$. In other words: if a configuration $c$ determines $x$, then the configuration $y\cdot c$ determines $yxy^{-1}$.
A: In an abelian group every element commutes with every other element. In non-abelian groups every element $g$ definitely commutes with all its powers $g^n$ (this included its inverse, and the identity element), other than that we cannot say without having specific info about the group.
A measure of failure of non-abelianness  is possible. We can ask the question how much an element $g$ 'contributes' towards non-abelinness?
If it commutes with an $x$, then $xgx^{-1}$ would be the same as $g$, otherwise an element different from $g$.
For a given $g$ try to list all the elements  $xgx^{-1}$, by varying $x$; the larger this list, more non-abelian the element $g$ is.  This is the conjugacy class of $g$.
