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Post Closed as "Not suitable for this site" by Derek Holt, Alex M., Carl-Fredrik Nyberg Brodda, coudy, LeechLattice
Proofreading
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LSpice
LSpice
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On thea quotient group $G=p^{n+m}.Q$ of a finite extension group. $G=p^{n+m}.Q$

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 equivalentOr equivalently that $G \cong p^n{{}^\cdot}(p^m{.}Q)$ (non-split extension of $p^n$ by $p^m{.}Q$?

On the quotient group $G=p^{n+m}.Q$ of a finite extension group. $G=p^{n+m}.Q$

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 equivalent 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 $G=p^{n+m}.Q$

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 $p^m{.}Q$?

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Isaac
Isaac
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On the quotient group $G=p^{n+m}.Q$ of a finite extension group. $G=p^{n+m}.Q$

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 equivalent that $G \cong p^n{{}^\cdot}(p^m{.}Q)$ (non-split extension of $p^n$ by $p^m{.}Q$?