Let $G=p^{n+m}.Q$ be an extension group of the special $p$-group $p^{n+m}$ by a group $Q$. Now $p^{n+m}=p^n{{}^\cdot}p^m$. How does one show that $\frac{G}{p^n}\cong p^m.Q$? Or equivalently that $G \cong p^n{{}^\cdot}(p^m{.}Q)$ (non-split extension of $p^n$ by $p^m{.}Q$?



On a quotient of a finite extension group $G=p^{n+m}.Q$
I presume you mean that $G$ has a normal subgroup $P$ with $G/P \cong Q$, where $P$ is a special $p$-group with $Z(P) = [P,P] = \Phi(P)$ elementary abelian of order $p^n$, and $P/Z(P)$ elementary abelian of order $p^m$.


But $Z(P)$ is characteristic in $P$ and hence normal in $G$, so $G$ has the structure ${p^n}^. (p^m.Q)$, and $G/Z(P)$ has structure $p^m.Q$.


