# If size of each conjugacy class is atmost $2$ then $G'\leq Z(G)$

Question is :

Show that if the size of each conjugacy class of a group $G$ is at most $2$ Then $G'\subseteq Z(G)$.

Suppose size of conjugacy class of an element $g\in G$ is $1$ i.e., $ngn^{-1}=g$ for all $n\in G$ i.e., $ngn^{-1}g^{-1}=1\in Z(G)$. So, $ngn^{-1}g^{-1}\in Z(G)$ for all $n\in G$ and $g\in G$ with conjugacy class having only one element.

i am stuck with the case of elements having conjugacy class order 2.

I would be thankful if someone can give hint for this.

Hint: The size of the conjugacy class of $x$ in $G$ is equal to $[G : C_G(x)]$.
• @TobiasKildetoft: Either $[G : C_G(x)] = 1$ or $[G : C_G(x)] = 2$. In both cases you can prove that $G' \leq C_G(x)$, so $G'$ centralizes every element of $G$. It helps to remember that subgroups of index $2$ are normal – Mikko Korhonen Aug 30 '13 at 11:18
• @mikko korhonen : I see that size of conjugacy class of $x\in G$ is $[G : C_G(x)]$ i.e., $C_G(x)$ is normal.. and then?? – user87543 Aug 30 '13 at 11:19
• @MikkoKorhonen : I have written my comment before reading yours... :). I see that $C_G(x)$ is normal but why does that mply $G'\leq C_G(x)$ – user87543 Aug 30 '13 at 11:21
• @PraphullaKoushik: If $G/N$ is abelian, then $G' \leq N$. Groups of order $2$ are abelian. – Mikko Korhonen Aug 30 '13 at 11:23