Conjugacy classes of the nonabelian group of order 21 How would you go about finding the conjugacy classes of the nonabelian group of order 21, $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$?

If $G$ is a nonabelian group of order $21$, then $G$ has trivial center. Otherwise $G/Z(G)$ would be cyclic and $G$ would be abelian.
Thus any element of order $3$ has its centralizer of order $3$ and thus has $7$ elements in its conjugacy class. By the same argument, an element of order $7$ has $3$ elements in its conjugacy class.
Let $a$ and $b$ be the number of conjugacy classes of order $3$ and $7$, respectively. By the class equation, $21 = 1 + 7a + 3b$. This implies that $a = b = 2$, because $a$ and $b$ are $\geq 1$ by Cauchy's theorem. Therefore there are five conjugacy classes: one for the identity, two containing elements of order $3$ and two containing elements of order $7$.
Since $y^{-1}xy = x^2$, we get $y^{-2}xy^2 = y^{-1}x^2y = x^4$. Therefore the conjugacy class of $x$ is $\{x, x^2, x^4\}$. The rest of the elements of order $7$ must be in the other conjugacy class, which is $\{x^3, x^5, x^6\}$.
We notice that $xyx^{-1} = yx$, $x^2yx^{-2} = yx^2$ and in general $x^jyx^{-j} = yx^{j}$. Thus in the two remaining conjugacy classes, one of them has all the elements of the form $yx^j$ and the other one all the elements of the form $y^2x^j$.
