Which finite groups have uniqueness for the ordered sequence of composition factors (up to isomorphism)?



Composition Series
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 $A\_0=1 < A\_1 < A\_2 < ... < A\_n = G$ and $B\_0=1 < B\_1 < B\_2 < ... < B\_n=G$. 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 $B\_{j-1}/K$. 


