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

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The group has a normal Sylow $7$-subgroup, generated by $x$, and it is clear from the way $y$ acts on $x$ that the conjugacy relation is generated by $x^i\sim x^{2i}$: this gives two conjugacy classes of elements of order $7$.

Using the Sylow theorems for $p=3$ and the fact that a Sylow $3$-subgroup cannot be normal (for otherwise the group would be abelian) you see that there is $1$ conjugacy class of $7$ cyclic subgroups of order $3$. They must be simply transitively permuted by the Sylow $7$-subgroup, so they give us 2 more conjugacy classes of elements of order three.

Finally, there's the class of $1$. In all, there are five classes then, which we can describe as follows: the classes of $1$, $y$, $y^2$, $x$, $x^3$.

We can check with GAP:

GAP4, Version: 4.4.10 of 02-Oct-2007, i686-pc-linux-gnu-gcc
Components:  small 2.1, small2 2.0, small3 2.0, small4 1.0, small5 1.0, small6 1.0, small7 1.0, small8 1.0, 
             small9 1.0, small10 0.2, id2 3.0, id3 2.1, id4 1.0, id5 1.0, id6 1.0, id9 1.0, id10 0.1, trans 1.0, 
             prim 2.1  loaded.
Packages:    AClib 1.1, Polycyclic 2.2, Alnuth 2.2.5, CrystCat 1.1.2, Cryst 4.1.5, Carat 2.0.2, AutPGrp 1.2, 
             CRISP 1.3.2, CTblLib 1.1.3, TomLib 1.1.2, FactInt 1.5.2, GAPDoc 1.2, IO 2.3, FGA, 
             IRREDSOL 1.1.2, LAGUNA 3.4, Sophus 1.23, Polenta 1.2.7, ResClasses 2.5.3, EDIM 1.2.3  loaded.
gap> List(AllSmallGroups(21, IsAbelian, false), g -> Length(ConjugacyClasses(g)));
[ 5 ]
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You don't mean "7 conjugacy classes of subgroups of order 3" Mariano, you mean "there are $7$ subgroups of order $3$, all of which are conjugate", as is clear from what you write later. A little argument is necessary to see that no element f ord $3$ is conjugate to its inverse (but in a group of odd order, no non-identity element is conjugate to its inverse). – Geoff Robinson Mar 7 '12 at 13:16
Yeah, that was a typo. As for your second observation: indeed, one has to prove that. But since there are 7 cyclic subgroups of order 3, which intersect trivially in pairs and each one of which is of course normalized by itself, the non-identity elements of each of them has a conjugacy class of size at most 7 . – Mariano Suárez-Alvarez Mar 7 '12 at 16:55

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 $y^2x^j$.

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Minor correction, the second conjugacy class of order 3 should be $\{x^3,x^5,x^6\}$. – Ruian Chen Apr 23 '15 at 16:58
@RuianChen: Thanks, fixed. – Mikko Korhonen Apr 24 '15 at 7:40

We can in fact generalize this situation as follows: Please see this question for reference

Let A be a normal subgroup of G such that A is the centralizer of every non-trivial element in A. If further G/A is abelian, then G has |G:A| linear characters, and (|A|−1)/|G:A| non-linear irreducible characters of degree =|G:A| which vanish off A.

Notice that the non-abelian group of order=$pq$ with $q \equiv 1 \mod{p}$ satisfies the conditions there, while that subgroup$A$ is given by a normal subgroup of order=$q$.
As when I post there is no answer which deploits characters, I post for the sake of references.
Thanks for paying attention.

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Here order of the group is $21= 3 \times 7$ . It is easy to see $3$ divides $7-1=6$. Thus, we have (upto isomorphism)two groups of order $21$. One of them is cyclic $\mathbb{Z}_{21}$ and other is non-abelian. This non-abelian is generated by two elements say $a$ and $b$ such that $|a|=3$ and $|b|=7$ and $ba=ab^r$ where $r$ is not congruent to $1$ modulo $7$ and $r^3\equiv 1 (mod 7)$. Defining conjugacy it is easy to see that there are five congugate classses and center has only the identity element. Moreover, class equation is $1+3+3+7+7\ldots$ :)

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