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My professor was presenting the problem that a group $G$ that has 3 conjugacy classes has either order 3 or 6. In his proof he said it was "clear" that by the class equation $$\begin{eqnarray} |G|&=&|Z(g)|+\sum |G:C_G(g_i)|\\ &=&1+|G:C_G(g_1)|+|G:C_G(g_2)| \end{eqnarray}$$ where the last two terms are given by the other two conjugacy classes that are not contained in the center. Then he proceeds from there to show that $|G:C_G(g_1)| \left(1+|G:C_G(g_2)|\right) $ and $|G:C_G(g_2)| \left(1+|G:C_G(g_1)|\right) $ and asked us to finish the rest of the proof. I'm not quite clear on the second equality above or how to proceed from here. Any help would be great. Thanks!

Edit: Here is the full statement. Let $G$ be a group of order $n$ finite. Suppose that $G$ has 3 congruency classes. Then the order of $G$ is either $3$ or $6$ and hence isomorphic to $\mathbb{Z}/3\mathbb{Z}$ or $S_3$.

user23793
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  • not clear please take a look again and edit it, what are you trying to say here, $|G:C_G(g_1)| \left(1+|G:C_G(g_2)|\right)$ and $|G:C_G(g_2)| \left(1+|G:C_G(g_1)|\right)$. You missed something after this line. Are they equal, or what? – Bhaskar Vashishth Oct 02 '14 at 04:07
  • I'm not sure what you mean. The line after this was left as an exercise and that's where I'm stuck. – user23793 Oct 02 '14 at 04:12
  • you are saying to show that this and this.....? fill the dots. this and this ..what? – Bhaskar Vashishth Oct 02 '14 at 04:15
  • take a look here http://math.stackexchange.com/questions/52350/finite-groups-with-exactly-n-conjugacy-classes-n-2-3 – Bhaskar Vashishth Oct 02 '14 at 04:18
  • I gave the full statement above. I'm not sure how else to explain the problem. – user23793 Oct 02 '14 at 04:19
  • My main question is why does the center have order 1. – user23793 Oct 02 '14 at 04:20
  • user23793: @BhaskarVashishth has answered your main question (why the center has order 1) in his answer below. In general it is a good idea to wait a little while before accepting an answer (and you should not accept an answer if you feel it does not answer your question) – zcn Oct 02 '14 at 05:33
  • @user23793 well atleast upvote it, it took efforts, some reputation must be awarded. thanks for acknowledging it – Bhaskar Vashishth Oct 02 '14 at 07:04
  • I'm sorry I was away from my computer. Thank you for the aid, It's helped me understand this a little bit more. – user23793 Oct 03 '14 at 01:03

2 Answers2

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Let $g=|G|$ and assume that we have $$g=1+a+b$$ in the obvious notation, with $a|g$ and $b|g$. Then $a|b+1$ and so $a\leq b+1$. Also $b|a+1$ so $b\leq a+1$ or $b-1\leq a$ thus $$b-1\leq a \leq b+1$$

So there are two possiblities (taking the symmetry of $a$ and $b$ into account) , $a=b$ and $a=b+1$

If $a=b$ then $1+2a=g$ so $a|1$ and $a=1$ and $g=3$.

If $a=b+1$ then $2+2b=g$ and $b|2$.

Now $b=1$ gives $a=2$ and $g=4$ but all groups of order $4$ are Abelian and have four classes.

So $b=2$ and $g=6$.

  • That's pretty clear, thank you. My last question though is why can we assume that $g=1+a+b$? Why can't the center have more elements than $1_G$ in it? Or am I not looking at this correctly? – user23793 Oct 02 '14 at 04:58
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OK. note that we are proving that $Z(G)$ is trivial only for non abelian cases. so let $G$ is non abelian and let $Z(G)$ is non trivial , so if $x \in Z(G)$ s.t. $x \neq 1$ then conjugacy class of $x$ is {$x$} as $gxg^{-1}=x$, so if let |$Z(G)$|=$2$, then two conjugacy classes are {$1$} and {$x$}, so let $g \in $ third conjugacy class, say $X$, so by class equation you have $|G|= 1+1+|G|/|C_G(g)|$ implies $1=2/|G|+1/|C_G(g)|$ which implies |$C_G(g)$|=$|G|/(|G|-2)$ , but we know $C_G(g)$ is a subgroup of $G$ so |$C_G(g)$| must givide |$G$| i.e. $|G|/(|G|-2)$ divides $|G|$, which is only possible for |$G$|=$3 , 4$, but groups of order $3,4$ are abelian. contradiction and so, $Z(G)$ is trivial for non abelian group with three conjugacy classes.

The case $Z(G)=3$ will give you |$G$|=$3$ by class equation and for greater than $3$ is not possible as only $3$ conjugacy classes are given

I hope I am clear now.

Rest i hope you have proved that then either $G$ is $\mathbb{Z_3}$ or $S_3$

Bhaskar Vashishth
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