Irreducible representation

Equivalently, $V$ is a simple $k[G]$-module (compare simple module).

Irreducible representations are the building blocks of completely reducible ones, and their characters are the irreducible characters.

Structural consequence (Schur)

A key fact is Schur's lemma: for irreducible $V$, $G$-equivariant endomorphisms of $V$ are very restricted (over an algebraically closed field, they are just scalars).

1. Cyclic groups over $\mathbb{C}$: all irreducibles are 1-dimensional. For $C_n=\langle g\mid g^n=1\rangle$ and $k=\mathbb{C}$, every irreducible representation is
   $$\rho_j(g)=\zeta_n^j\in \mathbb{C}^\times,\quad j\in\{0,1,\dots,n-1\},$$
   hence $\dim=1$. (These are precisely the complex characters of $C_n$.)

1. Irreducibles of $S_3$ over $\mathbb{C}$. Over $\mathbb{C}$, the group $S_3$ has exactly three irreducible representations up to isomorphism:

• the trivial 1-dimensional representation,
• the sign 1-dimensional representation $\mathrm{sgn}$,
• a 2-dimensional “standard” representation, realized as the action of $S_3$ on
  $$W_{\mathrm{std}}=\{(x_1,x_2,x_3)\in \mathbb{C}^3:\ x_1+x_2+x_3=0\},$$
  where $S_3$ permutes coordinates of $\mathbb{C}^3$.

1. Dependence on the field: a representation irreducible over $\mathbb{R}$ but reducible over $\mathbb{C}$. Let $G=C_3=\langle g\rangle$. Consider $V=\mathbb{R}^2$ with
   $$\rho(g)=\begin{pmatrix} \cos(2\pi/3)&-\sin(2\pi/3)\\ \sin(2\pi/3)&\cos(2\pi/3) \end{pmatrix}.$$
   This rotation has no real eigenvectors, hence no $1$-dimensional $G$-invariant subspace; therefore $V$ is irreducible over $\mathbb{R}$. After extending scalars to $\mathbb{C}$, $\rho(g)$ becomes diagonalizable with eigenvalues $e^{\pm 2\pi i/3}$, so $V\otimes_{\mathbb{R}}\mathbb{C}$ splits as a direct sum of two $1$-dimensional $\mathbb{C}$-representations.

See also: character orthogonality, number of irreducibles and conjugacy classes.