The following steps lead to a solution. The argument I am giving here is essentially equivalent to the argument given by Qiaochu above (in the comments) but is more general in that it is also applicable for compact Lie groups. The main point in this connection is in the application of the Peter-Weyl theorem in the solution to Exercise 4 below for compact Lie groups.
Let $G$ be a group and let $G'$ denote the commutator subgroup of $G$. We wish to understand how $G'$ acts on one-dimensional representations of $G$:
Exercise 1: Let $(\pi,V)$ be a one-dimensional representation of $G$. Prove that the induced representation of $G'$ is trivial.
The following result is fundamental in representation theory and is applicable to finite groups or, more generally, compact Lie groups:
Exercise 2: Prove that every finite-dimensional representation of $G$ is a direct sum of irreducible representations.
We know that every irreducible representation of $G$ is one-dimensional by assumption. In particular, every finite-dimensional representation of $G$ is a direct sum of one-dimensional representations.
Exercise 3 Prove that $G'$ acts trivially on every finite-dimensional representation of $G$.
Let us recall that a representation $(\pi,V)$ of $G$ is said to be faithful if the kernel $\text{ker }\pi$ is the trivial subgroup of $G$. The following exercise might be a little difficult; you can assume it on faith if desired:
Exercise 4 Prove that if $G$ is a finite group or a compact Lie group, then $G$ possesses a finite-dimensional faithful representation. (Hint: If $G$ is finite, then you can prove this using a familiar representation of $G$. If $G$ is an arbitrary compact Lie group, then appeal to the Peter-Weyl theorem if you can.)
The result should now be clear:
Exercise 5: Prove that if every irreducible representation of $G$ is one-dimensional, then $G'=\{e\}$, the trivial subgroup of $G$, i.e., $G$ is abelian.
I hope this helps!
Let us sort out things a bit.
Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation of any group $G$.
You have seen that if $\varphi : V \to V$ commutes with all $\rho(g)$, for $g \in G$,
that is
$$
\varphi (\rho(g) v) = \rho(g) \varphi (v)\tag{comm}
$$
for all $g \in G$ and $v \in V$,
then $\varphi = \lambda \operatorname{id}_{V}$ for some $\lambda \in \mathbf{C}$.
Now if $G$ is abelian we have $\rho(x) \rho(g) = \rho(g) \rho(x)$ for all $g, x \in G$, so that $\varphi = \rho(x)$ satisfies (comm).
It follows that for any $x \in G$ there is $\lambda \in \mathbf{C}$ such that
$$
\rho(x) = \lambda \operatorname{id}_{V},
$$
so all $\rho(x)$ are scalars, and then leave every subspace invariant.
Since the representation is irreducible, $V$ must have dimension $1$ then.
Best Answer
Take an irreducible complex representation (of dimension 1) and think of it as a real representation (of dimension 2). Clearly it either decomposes as either a sum of two one dimensional irreducibles or it is a two dimensional irreducible real representation.
But now I claim that every real representation arises this way: We know that every representation of $G$ over $\mathbb{R}$ or $\mathbb{C}$ appears inside the corresponding regular representation $\mathbb{R}[G]$ or $\mathbb{C}[G]$, but as a real representation $\mathbb{C}[G] = \mathbb{R}[G] + i\mathbb{R}[G]$ so every irreducible real representation appears inside the restriction of an irreducible complex representation.