If the product $x_1x_2...x_n=1$, prove that each $x_i=1$. Let $G$ be a finite solvable group and let $x_1,x_2,...,x_n$ be elements of $G$ of pairwise relatively prime orders. If the product $x_1x_2...x_n=1$, prove that each $x_i=1$.
I have no idea. Tell me some hints.
Thanks in advanced.
 A: First, you need to prove this is true if $G$ is abelian. This is simple, because in an abelian group raising a product to a power is the same as raising each element to that power. Since the order is relatively prime you can raise to the correct power to cancel out all but 1 element, forcing that element to be identity.
Now if $G/N$ is abelian where $N$ is a normal group, then using homomorphism from $G$ to $G/N$ to show that the image of each $x_{i}$ is identity (the order in the quotient group are still coprime). This force all $x_{i}$ to be in the kernel, ie. $N$.
Since the group is solvable, keep going down the chain till you each the trivial group.
A: This can be done by induction on the derived length of solvable group. If $G$ is abelian, the result is obvious. Now suppose that the derived length of $G$ is $m>1$. Consider $\bar{G}=G/G'$, where $G'$ is the derived subgroup of $G$. Let $\bar{x}_i=x_iG'$. Hence we get $\bar{x}_1\cdots \bar{x}_n=1$. By the abelian case, we see that $\bar{x}_i=1$. Hence we see that $x_i \in G'$ and the equation $x_1\cdots x_n=1$ holds in group $G'$, which has derived length $m-1$. Now by induction, $x_i=1$.
