most recent 30 from http://mathoverflow.net 2013-06-20T06:38:18Z http://mathoverflow.net/feeds/question/47695 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47695/centers-of-semidirect-products Greg Gibson 2010-11-29T16:21:50Z 2010-11-29T16:50:02Z

<p>The following question is for my own curiosity as I take some time to get reacquainted with group theory. </p> <p>Let G be a semi-direct product of the groups N and K with multiplication defined by the automorphism $\phi$ from K to Aut(N). Let Fix($\phi$) be the set of all elements of N that are mapped to themselves by all elements of the range of $\phi$. Clearly every element of Fix($\phi$) commutes with all elements of K and every element of the kernel of $\phi$ commutes with every element of N. </p> <p>If Fix($\phi$) is the trivial group in N and Ker($\phi$) is the trivial group in K, does that imply that the center of G is trivial? If so, could someone point me to a reference or proof. If not, then a counter example. </p>

http://mathoverflow.net/questions/47695/centers-of-semidirect-products/47698#47698 Robin Chapman 2010-11-29T16:50:02Z 2010-11-29T16:50:02Z

<p>Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$.</p> <p>As $y$ commutes with all elements of $N$ then $y$ lies in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$ is trivial.</p>