Intuition behind dividing groups into classes

Question

In group theory, a group $G$ can often be divided into classes. Two elements $U,V \in G$ belong to the same class if $\exists R\in G: RUR^{-1}=V$. This always strikes me as counter-intuitive. Apart from groups like symmetry groups, where one can assign "physical meaning" to the group elements, I lack understanding/a reason for why this distinction of classes is as powerful as it is. Why would an abstract group be divisible into classes? (Maybe a historical argument can help.)

Of course, I do not doubt that it is possible to do this when looking at concrete group tables and figuring out the classes manually. Is there an underlying reason for why dividing the group is most of the time possible with classes that do not only consist of one or all elements of the group?

Another thing is, why can an element only belong to exactly one class? The definition seems a bit lax; I sometimes wonder why there cannot be to different $R_1,R_2$ that sort some element into two different classes? Because then that is actually the same class for both elements...

Edit

This also comes from the fact that, when trying to devide a group into classes given only the group table and no "visual representation" (e.g. symmetry operations) for the group elements, how can at some point I know that I have found all elements of the class; that is, without calculating all possible combinations $RUR^{-1}$? "There might be some other combination of elements, that connects two classes..."

Answer 1

First of all it is important to understand why such classes exist. So let $G$ be a group. We define a relation on $G$ by setting $g_1\sim g_2$ if and only if there is $w\in G$ such that $wg_1w^{-1}=g_2$. It is easy to show that $\sim$ is an equivalence relation (try to prove this yourself). Therefore $G$ is "divided" into its equivalence classes (recall that an equivalence relation is the same as a partition).

To answer concretely your question, let's have a look at some particular examples

$\bullet$ $G$ is abelian: then for all $g_1,g_2$ and for all $w\in G$, $wg_1w^{-1}=g_1$ and therefore $g_1\sim g_2$ if and only if $g_1=g_2$. Therefore in this case the classes are all the $\{g\}$ for $g\in G$.

$\bullet$ If $G=S_n$ then it is a bit more tricky to find precisely the classes (two elements are equivalent if and only if they have the same cycle structure), but easy to see that, say, $(1)$ is not equivalent to $(12)$ (so there are nontrivial classes).

Now for every group, the only element equivalent to $1$, is $1$ since $w1w^{-1}=1$ implies that $w=1$. Therefore it cannot be that for some group $G$, $G$ is a class.

I hope this helps.

Answer 2

The fact that we are dealing with groups, rather than with amorphous sets, gives us the chance to define equivalence relations among their elements by means of group operation. The relation that you mention (conjugacy) is just one we can think of, of this kind; and I guess it is important ultimately because, given any two elements $a,b \in G$, the basic building blocks, namely the products $ab$ and $ba$, are conjugate (e.g. by $w=a^{-1}$); in fact: $ba=(a^{-1})(ab)(a^{-1})^{-1}$.

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Addendum

Perhaps, by looking at conjugacy relation explicitly in action may help.

For a given $a \in G$, call $\operatorname{Cl}(a):=\{gag^{-1}, g \in G\}$, namely $\operatorname{Cl}(a)$ is the set of all the elements of $G$ conjugate with $a$. Note that:

\begin{alignat}{1} c \in \operatorname{Cl}(a) \cap \operatorname{Cl}(b) &\iff c \in \operatorname{Cl}(a) \wedge c \in \operatorname{Cl}(b)\\ &\iff \exists g',g'' \in G \mid g'ag'^{-1}=g''bg''^{-1} \\ &\iff \exists g',g'' \in G \mid b=g''^{-1}g'ag'^{-1}g''=(g''^{-1}g')a(g''^{-1}g')^{-1} \\ &\iff \exists g''' \in G \mid b=g'''ag'''^{-1} \\ &\iff b \in \operatorname{Cl}(a) \\ \end{alignat}

whence:

$$b \notin \operatorname{Cl}(a) \Longrightarrow \operatorname{Cl}(a) \cap \operatorname{Cl}(b)=\emptyset \tag 1$$

Moreover, if $b \in \operatorname{Cl}(a)$, then $\exists \bar g \in G \mid b=\bar g a \bar g^{-1}$, whence:

\begin{alignat}{1} \operatorname{Cl}(a) &= \{gag^{-1}, g \in G\}\\ &= \{g(\bar g^{-1}b\bar g)g^{-1}, g \in G\} \\ &= \{(g\bar g^{-1})b(g\bar g^{-1})^{-1}, g \in G\} \\ &= \{g'bg'^{-1}, g' \in G\} \\ &= \operatorname{Cl}(b) \\ \tag 2 \end{alignat}

(the last but one equality holds because the map $G \to G$, defined by $g \mapsto g\bar g^{-1}$, is -in particular- surjective.) Lastly:

$$\forall a \in G, a\in\operatorname{Cl}(a) \tag 3$$

By $(1)$, $(2)$ and $(3)$, we have that $G$ is covered by disjoint subsets, or -equivalently- it is partitioned into (equivalence) classes. Each class is made of the elements of $G$ that are mutually conjugated

Answer 3

It seems you want to focus specifically on conjugacy classes. Let us start with two important and telling examples.

If $G=S_n$ is the symmetric group on $n$ elements, then for any $\sigma,\tau\in G$, $\sigma\tau\sigma^{-1}$ can be described as "the same as $\tau$, but with elements relabeled according to $\sigma$". For instance, if $\sigma$ is the permutation $(1\, 2)$, and $\tau=(134)(25)$, then $\sigma\tau\sigma^{-1}=(234)(15)$ (I rewrote $\tau$, but switched $1$ and $2$, which is what $\sigma$ does). Following this, you can see that two permutation are conjugate if and only if they are "the same up to some reordering on the elements". This shows why this relation makes a lot of sense: maybe you start with an actual permutation of concrete objects, then you assign an arbitrary number to each object, and this gives you a permutation in $S_n$; but you could have assigned different numbers, and you would have gotten any permutation conjugate to the initial one.

If $G=GL_n(K)$ is the group of invertible matrices over some field $K$, then for any $A,B\in G$, $ABA^{-1}$ can be described as "the same as $B$, but after the change of basis given by $A$". This gives the same kind of situation as for symmetric groups: two elements are in the same conjugacy class if they are the same up to a change of basis. Maybe you started with some non-singular linear map $V\to V$, and then chose a basis of $V$ to get an element of $GL_n(K)$; if you had chosen a different basis, you would have gotten any matrix in the same conjugacy class.

In general, a reasonable viewpoint is that elements in the same conjugacy class should be considered "the same from the point of view of $G$ itself". Although they are not literally equal, they must have the same algebraic properties. For instance, a very basic property of an element of a group is its order: the smallest integer $n$ such that $g^n=1$ (when it exists). In general, elements in a group can have very different orders, but elements in the same conjugacy class have the same order.

There is a little subtlety which is a little difficult to explain to someone with little experience in group theory: the difference between elements which are in the same conjugacy class (so they are the same up to an inner automorphism), and elements which are the same up to an arbitrary automorphism. I will try to give a somewhat convincing argument that conjugacy classes are really the correct ones in general (the difference does not come up for $GL_n$ and for $S_n$, except when $n=6$ for some weird reason).

Suppose $G$ is the group of symmetries of some structure $X$ (vector space, set, ring, topological space, whatever you want). Suppose you have another structure $Y$ of the same kind, and you know that actually they have the same structure (they are isomorphic). You may then choose some identification $f:X\to Y$ preserving the structure, and use $f$ to identify $G$ with the symmetry group of $Y$. So any symmetry (automorphism) $\sigma$ of $X$ corresponds to some symmetry of $Y$. But this depends on the choice of $f$! And if you chose a different $f$, then $\sigma$ will correspond to a different symmetry of $Y$ but... in the same conjugacy class.

So if you interpret $G$ as the symmetries of some object $X$, then conjugate elements are the same up to a change of point of view on $X$ (the two examples I gave at the beginning should make that statement clearer, I hope).