showing that $G$ is not solvable.

Question

suppose $G$ is a finite group and $1\neq a \in G$ ,$1\neq b \in G$ and $O(a)$ ,$O(b)$ ,$O(ab)$ every two of them are relatively prime ,then $G$ is not solvable.

my Idea:I suppose $G$ is solvable and put $H=<a,b>$ and consider $\frac{H}{H^{'}}$ and we should make a contradiction .

now I can't make it complete,it will be great to guide me to make contradiction,thanks.

now if we can show that there is some $a$,$b$,$c$ which $ab=c$ then it is solved.how we are sure that alwayes we can find such $a$,$b$ and $c$?

Answer 1

You are correct that it suffices to look at the subgroup $H$ generated by $a$ and $b$. We will show that $H$ is perfect (i.e. $H'=H$).

To that end we will denote the images under the canonical epimorphism $H \rightarrow H/H'$ by $\bar{h}$.

1. $O(\bar{h})$ divides $O(h)$ for all $h \in H$. (Why?).
2. In an abelian group (such as $H/H'$) the Order of $xy$ divides the product of the orders of $x$ and $y$. (Why?)
3. Use and 1. and 2. to conclude that $a,b \in H'$ (which is equivalent to $\bar{a}=\bar{b}=\bar{1}$) and hence $H=H'$.