# Decomposition of a group into a Product

I was wondering regarding the next question I encountered during my current research:

Given a p-group $G$ of order $p^n$ that can be decomposed into the product $G=AB$ of a normal subgroup $A$ and an arbitrary subgroup $B$ , what can we say about the orders of $A$ and $B$ ? Lagrange's theorem tells us that $A,B$ must be also p-groups, but can we say something about the power of p that their order has?

What if $B$ was cyclic? Does it change the answer?

If we also know that the rank of $G$ is $k$ , will it help us to determine the rank of $A$ ?

In another direction, given such a decomposition, how unique is it ? i.e.- have we got any way to estimate the number of ways in which we can decompose an arbitrary p-group into such a product?

Hope you'll be able to help. Any reference will be greatfully acknowledged

Thanks !

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Maybe I'm missing something, but it seems to me that one cannot say much. For instance, one can consider the abelian $p$-groups of order $p^n$ given by $\mathbb{Z} /p^a \mathbb{Z} \times \mathbb{Z} /p^b \mathbb{Z}$ with $a+b=n$, so any power of $p$ can occur. – Francesco Polizzi Apr 27 '12 at 15:17
Thanks a lot !!! – jason mfash Apr 28 '12 at 7:18

In every finite $p$-group $G$ of order $p^n$ there is a normal subgroup $N_k$ of order $p^k$ for any $k\le n$. The proof is by induction on $k$. For $k=1$ take a cyclic subgroup of order $p$ in the center, for the step, take a cyclic central subgroup in $G/N_k$ and its preimage in $G$. So if you do not assume that $A\cap B=\{1\}$, one can't say anything specific about the orders. If you assume that $A\cap B=\{1\}$, i.e. your product is semi-direct, then some information (not much, though) can be deduced.
If $|A|=p$ and $A$ is normal, then $A$ is central because in a nilpotent group every non-trivial normal subgroup intersects the center non-trivially. In that case either $B=G$ or $A\cap B=\{1\}$ and $G=A\times B$.
If $|A|=p^k, k\ge 2$, then again you can take a central cyclic of order $p$ subgroup $T$ of $A$, and you will have $G/T=(A/T)(BT/T)$. This reduces your problem to the same problem for a smaller group $G/T$.
"do you mean that the power of p in their order must sum up to n". That is correct. But one can say a little more. For example if $|N|=p$, then $N$ is necessarily a central subgroup (in a nilpotent group every normal subgroup has a non-trivial intersection with the center). Hence if $G$ is a semidirect product of $N$ and $A$, $|N|=p$, then it is a direct product: $G=N\times A$. – Mark Sapir Apr 28 '12 at 7:33