Algebra, Problems For Fun (27)

**NonCommAlg** · July 2nd 2009, 10:09 AM

Let $G$ be a finite group of odd order and $1 \neq g \in G.$ Prove that $g$ is not conjugate to $g^{-1}.$

**Bruno J.** · July 2nd 2009, 01:08 PM

Suppose $g=ag^{-1}a^{-1}$ .......... (1)

Then

$g^{-1}=aga^{-1}$

Combining with (1) : $g=ag^{-1}a^{-1} = a(aga^{-1})a^{-1}=a^2ga^{-2}=a^2(a^2ga^{-2})a^{-2}=a^4ga^{-4}=...$

So $g=a^2ga^{-2}=a^4ga^{-4}=a^6ga^{-6}...$

But since $|G|$ is odd, $<a>=<a^2>$ . In particular we find $k$ such that $a^{2k}=a$ . This yields $g=a^{2k}ga^{-2k}=aga^{-1}$ , so $g,a$ commute and $g^{-1}=g\Rightarrow g^2=1 \Rightarrow g=1$ .

**NonCommAlg** · July 3rd 2009, 05:09 PM

Originally Posted by Bruno J.

Suppose $g=ag^{-1}a^{-1}$ .......... (1)

Then

$g^{-1}=aga^{-1}$

Combining with (1) : $g=ag^{-1}a^{-1} = a(aga^{-1})a^{-1}=a^2ga^{-2}=a^2(a^2ga^{-2})a^{-2}=a^4ga^{-4}=...$

So $g=a^2ga^{-2}=a^4ga^{-4}=a^6ga^{-6}...$

But since $|G|$ is odd, $<a>=<a^2>$ . In particular we find $k$ such that $a^{2k}=a$ . This yields $g=a^{2k}ga^{-2k}=aga^{-1}$ , so $g,a$ commute and $g^{-1}=g\Rightarrow g^2=1 \Rightarrow g=1$ .

you can also use induction to show that $g=a^ng^{-1}a^{-n}$ for all odd numbers $n$ and then choose $n=o(a)$ to get $g^2=1$ and thus $g=1.$

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