# Thread: product of n-1 commutators

1. ## product of n-1 commutators

I'm probably missing something very simple here, but I can't figure this one out.

Let $x_{1},...,x_{n}$ be elements in a group $G$.
show that an element of the form $x_{1}x_{2}...x_{n}x_{1}^{-1}x_{2}^{-1}...x_{n}^{-1}$ is a product of $n-1$ commutators.

I tried this with induction on $n$, but that led nowhere and I tried showing how an element of the form $abca^{-1}b^{-1}c^{-1}$ was a product of 2 commutators but i got stuck.

Any hint would be appreciated

SK

2. Originally Posted by skyking
I'm probably missing something very simple here, but I can't figure this one out.

Let $x_{1},...,x_{n}$ be elements in a group $G$.
show that an element of the form $x_{1}x_{2}...x_{n}x_{1}^{-1}x_{2}^{-1}...x_{n}^{-1}$ is a product of $n-1$ commutators.

I tried this with induction on $n$, but that led nowhere and I tried showing how an element of the form $abca^{-1}b^{-1}c^{-1}$ was a product of 2 commutators but i got stuck.

Any hint would be appreciated

SK

$abca^{-1}b^{-1}c^{-1}=[a,bc][b,c]$ (defining $[x,y]:=xyx^{-1}y^{-1}$ ) .

Tonio

3. Originally Posted by tonio
$abca^{-1}b^{-1}c^{-1}=[a,bc][b,c]$ (defining $[x,y]:=xyx^{-1}y^{-1}$ ) .

Tonio
OK, so this is what I get then:

$[x_{1},x_{2}...x_{n}][x_{2},x_{3}...x_{n}]...[x_{n-1},x_{n}]=x_{1}...x_{n}x_{1}^{-1}...x_{n}^{-1}$

Thanks for the hint,

SK