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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

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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.

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  • $\begingroup$ 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? $\endgroup$ – ZFR Oct 13 '18 at 21:38
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    $\begingroup$ 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. $\endgroup$ – Mark Oct 13 '18 at 21:41
  • $\begingroup$ If the powers are positive then OK, what if one of them is negative? $\endgroup$ – ZFR Oct 13 '18 at 21:48
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    $\begingroup$ 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. $\endgroup$ – Mark Oct 13 '18 at 21:51
  • $\begingroup$ Thanks a lot, dear Mark! You solutions are always elegant and detailed! $\endgroup$ – ZFR Oct 13 '18 at 23:26

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