# $K(G)=3 \Longrightarrow G\cong\mathbb Z_3\ \mathrm{or} \ G\cong S_3$

According to J.S. Rose book "A Course on Group Theory":

In class equation $$|G|=\sum_{i=1}^k|G:C_G(x_i)|$$ where $x_1,x_2,...,x_k\in G$ one from each of above $k$ classes; $K(G)$ is called the class number of $G$.

Now I want to verify:

$$K(G)=3 \Longrightarrow G\cong\mathbb Z_3\ \mathrm{or} \ G\cong S_3$$

If $K(G)=3$ then I see $|Z(G)|=1$, $|Z(G)|=2$ or $|Z(G)|=3$. $|Z(G)|=3$ leads $G$ to be abelian so I have $G\cong\mathbb Z_3$. If $|Z(G)|=2$ so I have an element, say $x$, in $G$ which doesn't belong to its center. Therefore $d=|G:C_G(x)|\big|\ |G|$ and so $d=1$ or $d=2$. It is clear to me that these two make contradictions. I see myself very close to $S_3$ when $|Z(G)|=1$. Please, if I am on a right way help me about the final choice $|Z(G)|=1$. Thanks

-
Have a look here: math.stackexchange.com/questions/52350/… –  Lucien Sep 6 '12 at 9:59
Or problem 6 on page 2 of math.cornell.edu/~slim/Prelim_II_4320_sol.pdf –  Gerry Myerson Sep 6 '12 at 13:17
@Lucien: Thanks for the link. –  Babak S. Sep 6 '12 at 13:52
@Babak Sorouh:you are welcome. –  Lucien Sep 7 '12 at 12:52

## 1 Answer

If $|Z(G)|=1$, then above class equation reduces to $1=\frac{1}{|G|}+ \frac{1}{|C_G(x)|}+\frac{1}{|C_G(y)|}$.

If $C_G(x)|,|C_G(y)|\geq 3$, then as $C_G(x)\lneq G$, the RHS is less than $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$, a contradiction.

Suppose $|C_G(x)|=2$. Now, $|G|=k.|C_G(y)|$ for some $k>1$. Then, above equation reduces to

$\frac{1}{2}=\frac{1}{k|C_G(y)|}+\frac{1}{|C_G(y)|}=\frac{k+1}{k}\frac{1}{|C_G(y)|}$; hence $|C_G(y)|=\frac{2(k+1)}{k}$ and it is integer; i.e. $k|2(k+1)$.

Now, if $k>2$, then $2k+2<2k+k=3k$; hence $2(k+1)\in \{k,2k\}$, both leads to contradiction.

Therefore, $k=2$, and $|C_G(y)|=3$.

By first equation, we deduce $|G|=6$. ....$q.e.d.$

-