Originally, I want to show that following statement.
Let $F$ be a field. $GL_n(F)$ has a composition series if and only if $F$ is finite.
Here is my approach. In this, I assumed that the existence of refinement of group that has a composition series.
If $F$ is finite, then $GL_n(F)$ is also finite so has a composition series. Suppose $F$ is infinite field but $GL_n(F)$ has a composition series. Consider the refinement of subnormal series $$ 1 \leq SL_n(F) \leq GL_n(F). $$ Let $$ SL_n(F) \leq N_1 \leq \cdots \leq N_m = GL_n(F) $$ is the part of the refinement. From this we have a subnormal series of $GL_n(F)/SL_n(F)$, that is, $$ SL_n(F)/SL_n(F) \leq N_1/SL_n(F) \leq \cdots \leq N_m/SL_n(F) = GL_n(F)/SL_n(F). $$ Note that this is also a composition series by the third isomorphism theorem. Since $GL_n(F)/SL_n(F)$ is isomorphic to $F^{\times}$, we conclude that $F^{\times}$ is also has a composition series, which is contradict to the assumption $F$ is infinite. Therefore $GL_n(F)$ cannot have a composition series.
So I want to know that my assumption is true, and if it is true, how prove it. I'm very happy if you let me know.
Moreover pointing out the errors in the proof in the 'box' will be welcome.
I think you need to note that an abelian group with a finite composition series is finite, by noting that simple abelian groups are cyclic of prime order. This justifies the final contradiction.
To justify passing to the quotient, suppose $G$ has a finite composition series $1 = H_0 \leq \cdots \leq H_r =G$ and $N$ is a normal subgroup of $G$. Consider the series $N=H_0 N \leq \cdots \leq H_r N = G$. Then each subquotient $H_i N/H_{i-1} N \cong H_i / H_{i} \cap H_{i-1}N$ is either trivial or simple because $H_{i-1} \leq H_i \cap H_{i-1}N \leq H_i$. Thus $G/N$ has a finite composition series as well.