What is the intuition of conjugacy classes?

Solution 1:

There is one way to understand conjugacy from the perspective of linear algebra, using change-of-basis.

Conjugacy can also be interpreted as a "measure of departure" from being a commutative group. Given two elements $g,h$ of a group $G$ let us look at the element $h^{-1}gh$.

We modify $g$ by multiplying by $h$ on the right. To get back the $g$ we should multiply this result by $h^{-1}$ again on the right. But instead we multiplied it by $h^{-1}$ on the left. That is we used the same ingredients "but mixed them in different order". If it gives back $g$ well and good then these elements are said to commute. If it does not give back $g$, we interpret it as something "close to $g$, a conjugate of $g$. In an abelian group if two elements are close enough to be conjuagate they are in fact same. In a non-abelain group we consider $h^{-1}gh$ for various elements $h$ as giving use many elements that are close to $g$, or 'similar to" $g$ (Linear algebra usus this terminology.)

If an element has many elements similar/conjugate to it (other than itself) it is a measure of failure of commutativity. The larger the number of elements conjugate to it, more non-commutative that element is.

Conjugate elements have same order, and have same behaviour in many respects. (In matrices conjugate elements have same trace, determinant and eigenvalues).

Solution 2:

It's not easy to see conjugacy in a group table, but try this:

Look at the $f_v$ row, and the $f_h$ column. $f_d$ is in both of them, and it's in the $r_1$ column of the $f_v$ row, and the $r_1$ row of the $f_h$ column. That tells you $f_vr_1=r_1f_h$ (since both equal $f_d$), which says $f_v=r_1f_hr_1^{-1}$, that is, $f_v$ and $f_h$ are conjugate.

In general, $a$ and $b$ are conjugate if there are elements $x$ and $y$ such that $x$ is in the $a$ row and the $y$ column, and also in the $b$ column and the $y$ row. If there are no such $x$ and $y$, then $a$ and $b$ aren't conjugate.

Of course, this is nothing but a restatement of the definition of conjugacy --- I'm just trying to restate it in terms of the group table, as requested.

Solution 3:

Let me use an analogy from linear algebra. In linear algebra two matrices $X, Y$ are conjugate iff. there exists a matrix $M$ such that $X = MYM^{-1}$. In linear algebra, this has a specific interpretation: $X$ is the change of basis of $Y$ where $M$ is the matrix representing the transformation of the basis of $Y$ to the basis of $X$.

It turns out that this interpretation transfers exactly into group theory. Let us consider the symmetry group $S_n$ and let us express the elements of $S_n$ in cycle notation. let $x = (abc)(de)$, $y = (cba)(de)$, and $g = (ca)$. The element $x$ can be considered as a "change of basis" (with the change being switching $c$ for $a$ $y$) from $y$ with the element representing the "basis transformation" (substitution in this case) being $g$.