# Thread: Group proof by induction

1. ## Group proof by induction

prove by induction that for elements x and g of a group,

(x^-1.g.x)^n=x^-1.g^n.x

My working.

I've done the easy parts, showing true for n=1, assuming true for n=k but now given that, I need to show true for n=k+1.

I worked with LHS, expanding the braket and trying to manpiulate to get an expression I already knew but I'm pretty stumped.

Thanks for any help.

2. Use:

$\displaystyle (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k$

Now, apply the induction hypothesis.

Fernando Revilla

3. Originally Posted by FernandoRevilla
Use:

$\displaystyle (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k$

Now, apply the induction hypothesis.

Fernando Revilla

so I obtain (x^-1.g^k.x)(x^-1g.x) but I can't assume commutativity so how do I simplify to x^-1.g^k+1.x.

4. You needn't apply commutativity:

$\displaystyle (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k=(x^{-1}gx)(x^{-1}g^kx)=\ldots=x^{-1}g^{k+1}x$

Fernando Revilla

5. Originally Posted by FernandoRevilla
You needn't apply commutativity:

$\displaystyle (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k=(x^{-1}gx)(x^{-1}g^kx)=\ldots=x^{-1}g^{k+1}x$

Fernando Revilla
can I just multiply the middle x and x^-1?

6. Originally Posted by Plague01
can I just multiply the middle x and x^-1?

Of course you can. Use the associative property and $\displaystyle xx^{-1}=1$ .

Fernando Revilla