# Thread: Algebra, Problems For Fun (27)

1. ## Algebra, Problems For Fun (27)

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

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

Then

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

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

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

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

3. Originally Posted by Bruno J.
Suppose $\displaystyle g=ag^{-1}a^{-1}$ .......... (1)

Then

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

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

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

But since $\displaystyle |G|$ is odd, $\displaystyle <a>=<a^2>$. In particular we find $\displaystyle k$ such that $\displaystyle a^{2k}=a$. This yields $\displaystyle g=a^{2k}ga^{-2k}=aga^{-1}$, so $\displaystyle g,a$ commute and $\displaystyle g^{-1}=g\Rightarrow g^2=1 \Rightarrow g=1$.
you can also use induction to show that $\displaystyle g=a^ng^{-1}a^{-n}$ for all odd numbers $\displaystyle n$ and then choose $\displaystyle n=o(a)$ to get $\displaystyle g^2=1$ and thus $\displaystyle g=1.$