Decomposition of a group into a Product - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:17:10Z http://mathoverflow.net/feeds/question/95366 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95366/decomposition-of-a-group-into-a-product Decomposition of a group into a Product jason mfash 2012-04-27T15:05:27Z 2012-05-05T14:11:05Z <p>I was wondering regarding the next question I encountered during my current research:</p> <p>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? </p> <p>What if $B$ was cyclic? Does it change the answer?</p> <p>If we also know that the rank of $G$ is $k$ , will it help us to determine the rank of $A$ ? </p> <p>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?</p> <p>Hope you'll be able to help. Any reference will be greatfully acknowledged</p> <p>Thanks ! </p> http://mathoverflow.net/questions/95366/decomposition-of-a-group-into-a-product/95373#95373 Answer by Mark Sapir for Decomposition of a group into a Product Mark Sapir 2012-04-27T16:24:10Z 2012-05-05T14:11:05Z <p>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. </p> <p>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 $G=A\times B$. </p> <p>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 $G/T$. </p>