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 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 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>