Let $G$ be a group having a composition series and $H$ a normal subgroup of $G$. Show that there exists a composition series in which one term is $H$.
My idea is to use the Schreier refinement theorem.
Let $G=G_0 \triangleright G_1 \triangleright G_2 \triangleright \dots \triangleright G_m = \{ e \}$ be a composition series for $G$. Now consider $G\triangleright H \triangleright \{ e\}$, if this is a composition series we are done.
If not, by the Schreier refinement theorem there exists equivalent refinements of these two series. Call these refinements $R_1$ and $R_2$, where $R_1$ is the refinement of the composition series. Now because these two refinements are equivalent the factors of $R_2$ are isomorphic to factors of $R_1$. Now because $R_1$ is a refinement of a composition series this means all of its factors are simple groups. Thus all the factors of $R_2$ are simple. Thus if we remove groups from $R_2$ in a way such that we eliminate trivial factors this should give us a composition series for $G$ that contains $H$.
Does this solution seem correct?
Thanks for the feedback!
Best Answer
I don't see why you want to remove factors of $R_{2}$. Anyway, you have $R_{2}$ which is a subnormal serie that contains H, where all the factors are simple as you said. There is a theorem which states that a subnormal serie is a composition serie iff all its factors are simple. So we have immediately that $R_2$ is a composition serie.
The proof of the the theorem used is easy: Let be $G \trianglerighteq G_{1} \trianglerighteq ... \trianglerighteq 1$. Its factors are $G_{i}/G_{i+1}$.
$G_{i}/G_{i+1}$ simple $\Leftrightarrow$ $\nexists H/G_{i} \trianglelefteq G_i/G_{i+1}$ $\Leftrightarrow$ $\nexists G_{i+1} \vartriangleleft H \vartriangleleft G_i $ which means that I can't make a proper refinement. For the second iff the third isomorphism theorem it's used.