Irreducible representation

Remarks

Example: let $G = \Z/4\Z$ act on $\R^2$, with the class of $1$ acting as the rotation $R$ by $90$ degrees. A $G$-invariant line would be spanned by an eigenvector of $R$, but $R$ has characteristic polynomial $x^2 + 1$ with no real root. So the only invariant subspaces are $0$ and $\R^2$, and the representation is irreducible. Non-example: let $G = \Z$ act on $\C^2$, with $n$ acting by the matrix $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$. The line spanned by $(1, 0)$ is invariant, so the representation is not irreducible. It is also not a direct sum of irreducible representations, because this line has no invariant complement. Over $\C$, the same rotation representation of $\Z/4\Z$ on $\C^2$ is not irreducible: it splits into the eigenlines for $i$ and $-i$. So irreducibility depends on the field. An irreducible representation is also called simple, since it is the same as a simple module over the group algebra.