If a group has a faithful reducible two-dimensional representation then its commutant is abelian. 
A group $G$ has a faithful reducible two-dimensional representation. Prove that commutant of the group $G'$ is Abelian.

I think to so.  Commutant $G'\triangleleft G$. Let $\rho$ is the faithful reducible two-dimensional representation of G. Since $\rho$ is the faithful representation of G, $\ker \rho =e$.
Any hints will be greatly appreciated. Sorry for my poor English.
 A: Let $V$ be the relevant $G$-module, and $F$ the underlying field. By assumption, there is a subspace $U$, with $\{0\} \ne U \ne V$ which is invariant under $G$. Take a non-zero vector of $U$ and add an element of $V \setminus U$ to get a basis of $V$.
With respect to this basis, every element of $G$ can be represented as a matrix of the form
$$
\begin{bmatrix}a & b\\0 & c\end{bmatrix}.
$$
So $G$ is a subgroup of the matrix group
$$
\mathfrak{G} = \left\{
\begin{bmatrix}a & b\\0 & c\end{bmatrix}
: a, b, c \in F
\right\}.
$$
Now it is a simple matter to show that
$$
G'\le \mathfrak{G}' = \left\{
\begin{bmatrix}1 & b\\0 & 1\end{bmatrix}
: b\in F
\right\},
$$
the latter being clearly an abelian group.
A: Must not $G$ itself be abelian? If we reason via character theory: let $\rho$ be a 2-dimensional faithful reducible character. Then $\rho=\lambda+\mu$, where $\lambda$ and $\mu$ are both linear characters. Since $ker(\rho)=1=ker(\lambda) \cap ker(\mu)$ and the kernel of a linear character always contains $G'$, we get $G'=1$, that is, $G$ must be abelian.
