Composition Series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:12:50Z http://mathoverflow.net/feeds/question/104911 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104911/composition-series Composition Series solovei 2012-08-17T13:00:50Z 2012-08-18T08:15:15Z <p>Which finite groups have uniqueness for the ordered sequence of composition factors (up to isomorphism)?</p> http://mathoverflow.net/questions/104911/composition-series/104947#104947 Answer by Mark Sapir for Composition Series Mark Sapir 2012-08-17T21:14:20Z 2012-08-18T08:15:15Z <p>Here is a characterization. A group \$G\$ has two different composition series if and only if it has a factor \$H/K\$ which is a direct product of two non-isomorphic simple subgroups, where \$H\$ is a subnormal subgroup of \$G\$, \$K\$ is a normal subgroup of \$H\$. Indeed, if such \$H/K\$ exists, then clearly there are two different composition series. Conversely, suppose that there are two different composition series \$A_0=1 &lt; A_1 &lt; A_2 &lt; ... &lt; A_n = G\$ and \$B_0=1 &lt; B_1 &lt; B_2 &lt; ... &lt; B_n=G\$. Let \$j\$ be the last index with \$A_{j-1}\ne B_{j-1}\$ (non-isomorphic), \$n\ge j\ge 1\$. Let \$H=A_j=B_j\$. Note that \$A_{j-1}\$ and \$B_{j-1}\$ are normal in \$H\$. Hence \$A_{j-1}B_{j-1}\$ is normal in \$H\$. Since it is bigger than \$A_{j-1}\$, we have \$H=A_{j-1}B_{j-1}\$. Hence \$K=A_{j-1}\cap B_{j-1}\$ is normal in \$H\$ and \$A_{i-1}/K\$ is isomorphic to \$H/B_{j-1}\$ hence simple. Similarly \$B_{j-1}/K\$ is simple. Now \$H/K\$ has two different normal simple subgroups \$A_{j-1}/K\$ and \$B_{j-1}/K\$ with trivial intersection, so \$A_{j-1}/K\$ and \$B_{j-1}/K\$ commute and form a direct product. Therefore \$H/K\$ is isomorphic to the direct product of two non-isomorphic simple groups \$A_{j-1}/K\$ and \$B_{j-1}/K\$. </p>