To summarize the discussion. Let $G$ be a subgroup of $\mathrm{SL}_2(\mathbf{C})$, $H$ its Zariski closure.

**Proposition.** Equivalent statements:

(i) $G$ acts irreducibly on $\mathbf{C}^2$;

(ii) $H$ acts irreducibly on $\mathbf{C}^2$;

(iii) $G$ fixes no point on $\mathbb{P}^1_\mathbf{C}$;

(iv) $H$ fixes no point on $\mathbb{P}^1_\mathbf{C}$;

(v) One of the following holds:

$\quad$(a) $H$ (and hence $G$) is finite and non-abelian;

$\quad$(b) $H$ is conjugate to the monomial group, made up of diagonal matrices and anti-diagonal matrices with determinant 1;

$\quad$(c) $H=\mathrm{SL}_2(\mathbf{C})$ (i.e., $G$ is Zariski-dense).

*Proof.* The equivalence between (i),(ii),(iii),(iv) is trivial (although that iii/iv implies i/ii is specific to dimension 2).

For a finite group in dimension 2 in characteristic zero, non-irreducibility implies that the action is diagonalizable; hence (a) implies (i). The only fixed points by the group of diagonal matrices with determinant 1 are the two coordinate axes; then are switched by the monomial group, hence (b) implies (iv), and (c) implies (iv) follows (and is clear anyway).

Conversely, suppose that none of (a),(b),(c) holds. If $G$ is finite, this means that $G$ is abelian, its irreducible representations have dimension 1 and the negation of (i) follows. Otherwise, we discuss on the Lie algebra of $H$. If it is conjugate to the upper unipotent or upper triangular subalgebra, then the corresponding connected group fixes a unique point, which is then fixed by $H$ and we obtain the negation of (ii). Since (c) does not hold, it is then conjugate to the subalgebra of diagonal matrices. Hence the Zariski connected component of $H$ consists of the group $D$ diagonal matrices with determinant 1; it has index two in its normalizer (monomial matrices). Since (b) does not hold, we deduce that $H=D$, and again is abelian and does not act irreducibly. $\Box$