0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

Hint: The size of the conjugacy class of $x$ in $G$ is equal to $[G : C_G(x)]$.

$\endgroup$
  • $\begingroup$ This was my first thought when I saw the question, and it did not directly lead me anywhere. Maybe you could elaborate on the hint? $\endgroup$ – Tobias Kildetoft Aug 30 '13 at 11:02
  • $\begingroup$ @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 $\endgroup$ – Mikko Korhonen Aug 30 '13 at 11:18
  • $\begingroup$ @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?? $\endgroup$ – user87543 Aug 30 '13 at 11:19
  • $\begingroup$ @MikkoKorhonen : I have written my comment before reading yours... :). I see that $C_G(x)$ is normal but why does that mply $G'\leq C_G(x)$ $\endgroup$ – user87543 Aug 30 '13 at 11:21
  • 1
    $\begingroup$ @PraphullaKoushik: If $G/N$ is abelian, then $G' \leq N$. Groups of order $2$ are abelian. $\endgroup$ – Mikko Korhonen Aug 30 '13 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy