MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 !

share|cite|improve this question
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 $G/T$.

share|cite|improve this answer
Thanks ! That was my first intuition indeed... When you say that if the product is semi-direct then we can say a bit more, do you mean that the power of p in their order must sum up to n? Thanks again ! – jason mfash Apr 28 '12 at 7:19
"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
Thanks a lot ! ! – jason mfash Apr 29 '12 at 6:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.