Let $G$ of order $p^{\alpha}$, (where $p$ is a prime ) be a group such that all of its subgroups are normal. Let $g$ be the lowest order element ($p^{\lambda})$ which is not self-conjugate in $G$. Let $h$ be any element of $G$ which is not permutable ($gh \neq hg$) with $g$ and let the order of $h$ is $p^{\mu} (\mu \ge \lambda)$ .
Since every subgroup of $G$ is normal, it follows that $h^{-1}gh$ is in group generated by $g$ i.e. $\{g\}$. Then $h^{-1}gh$ is a power of $g$. Write $g^{-}h^{-1}gh =c$. Then $c$ is a power of $g$. In a similar way it may be shown that $c$ is a power of $h$. Hence $c$ is in the greatest common subgroup $D$ of $\{g\}$ and $\{h\}$. Every subgroup of $\{g\}$ contains $\{g^{p^{\lambda -1}}\}$, and every subgroup of $\{h\}$ contains $\{h^{p^{\mu -1}}\}$ . But since, $c\neq 1$, $D$ does not consist of the identity alone. So $D$ contains both $\{g^{p^{\lambda -1}}\}$ and $\{h^{p^{\mu -1}}\}$. Therefore $g^{p^{\lambda -1}} = h^{p^{u (\mu -1)}}$, where $u$ is an appropriate integer which is co-prime to $p$. Since $g$ and $h$ are not permutable , it follows that $D$ does not concide with $\{g\}$. Therefore $\lambda $ greater than 1.
Question 1: Why they are picking the smallest order element $g$?
Question 2 : How $g^{p^{\lambda -1}} = h^{p^{u (\mu -1)}}$
Question 3: How therefore $\lambda $ greater than 1?
Yo can read from these two :
