# Simple nonabelian where orbits have length at least 3

If $$G$$ is a simple nonabelian group and $$e\neq x \in G$$, show that $$x$$ must have at least 3 conjugates (including itself).

My attempt: Suppose $$G$$ acts on itself by conjugation, i.e. for any $$g\in G$$, $$\pi_g:G\to G$$ by $$\pi_g(y)=gyg^{-1}$$ for all $$y\in G$$.

It is easy to see that $$x$$ is fixed point iff $$x\in Z_G$$, but since $$G$$ is nonabelian simple group then $$Z_G=\{e\}$$. Since $$x\neq e$$ then $$x$$ is not fixed point then $$\exists g_0\in G$$ such that $$\pi_{g_0}(x)\neq x$$ or $$g_0xg_0^{-1}\neq x$$. By definition $$\text{Orb}_x=\{gxg^{-1}:g\in G\}$$ and we need to show that $$\text{Orb}_x$$ has length at least three.

But we have shown that $$x, g_0xg_0^{-1}\in \text{Orb}_x$$. How to show that it has one more element?

Would be very grateful for hint

Suppose $$Orb_x=\{x,y\}$$. Then we know $$yxy^{-1}$$ must be either $$x$$ or $$y$$, because $$x$$ is not conjugate to any other elements. If $$yxy^{-1}=y$$ then we get $$xy^{-1}=e$$ and hence $$x=y$$ which is a contradiction. So it must be $$yxy^{-1}=x$$ and hence $$yx=xy$$. It follows that the subgroup $$H=\langle x,y\rangle\leq G$$ is abelian. Now I'll leave you to check that actually $$H\trianglelefteq G$$, and because $$G$$ is simple it implies $$H=G$$ and we get $$G$$ is abelian which is a contradiction.
• Thanks for reply but let me ask you a question: $H$ is a subgroup generated by $x$ and $y$, right? and we know that $xy=yx$ why it follows that $H$ is abelian? Could you clarify this moment, please? – ZFR Oct 13 '18 at 21:38
• Let $g,h\in H$. Because $xy=yx$ we can write them in the form $g=x^iy^j$ and $h=x^ky^l$. And then $gh=x^iy^jx^ky^l=x^ky^lx^iy^j=hg$ simply by using the fact that $x$ commutes with $y$, powers of $x$ of course commute with each other, as well as powers of $y$ commute with each other. – Mark Oct 13 '18 at 21:41
• It doesn't matter. Centralizer of an element is a group so it contains inverses. If $x$ commutes with $y$ then it commutes with $y^{-1}$ as well. You know that $xyx^{-1}=y$. Take the inverse on both sides and you will get $xy^{-1}x^{-1}=y^{-1}$ which implies $xy^{-1}=y^{-1}x$. In the same way you can show $x^{-1}$ and $y$ commute with each other, as well as $x^{-1}$ and $y^{-1}$ commute with each other. All of their powers commute. And that means if you have a product of powers of $x$ and powers of $y$ you can change the order as you wish. – Mark Oct 13 '18 at 21:51