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I have referred the question already asked , i understood almost all but this line below -

"I know it contains $N_{k}$ and some stuff not in $N_{k}$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal."

link - A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0\le k \le n$

Please help with the above theorem.

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BAYMAX
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  • I understood that $\pi^{-1}(C(G/N_k))$ is normal in $G$ , but what happens after that ? , how do we reach to conclude that $G$ contains a subgroup of orders of the form $p^{k}, 0 \leq k \leq n$? – BAYMAX Feb 14 '17 at 10:10

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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$The point is, $C(G/N_{k})$ has order divisible by $p$, hence it has a subgroup of order $p$. (You may also use induction, if you wish.)

By the correspondence theorem this subgroup has the form $H/N_{k}$, for some subgroup $H$ containing $N_{k}$.

Then $H$ has order $p^{k+1}$, as $$ p = \Size{\frac{H}{N_{k}}} = \frac{\Size{H}}{\Size{N_{k}}} = \frac{\Size{H}}{p^{k}}. $$

Finally, $H$ is normal in $G$ because, again by the correspondence theorem, $H/N_{k}$ is normal in $G/N_{k}$, as every subgroup of $C(G/N_{k})$ is normal in $G/N_{k}$.

Andreas Caranti
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