Let \(V\) be a vector space over some field \(k\) and let \(G\) be a subgroup of \(\mathrm{GL}(V)\). We say that \(G\) is irreducible if there are no \(G\)-invariant subspaces of \(V\) (other than \(0\) and \(V\)).

Example 1.5. We give a few examples of irreducible matrix groups.

1. (i) The groups \(\mathrm{GL}(V)\) and \(\mathrm{SL}(V)\) are irreducible matrix groups.
2. (ii) Thought of as rotations and reflections of a square, the group \(D_8\) forms a \(2\)-dimensional matrix group over any field of characteristic \(0\), and it is irreducible.
3. (iii) Similarly the rotations of a cube, which form the group \(S_4\), yield a \(3\)-dimensional irreducible matrix group over any field of characteristic \(0\).

Example 1.6. Instead of all rotations and reflections of a square, we consider the \(180^\circ\) rotation (and the identity) and two reflections. This gives a copy of the Klein four group \(V_4\).

If the reflections are in the \(x\)- and \(y\)-axes then the copy of \(V_4\) is reducible, with \(G\)-invariant subspaces the \(x\)- and \(y\)-axes. If the reflections are in the lines \(x=\pm y\), then the copy of \(V_4\) is again reducible, with \(G\)-invariant subspaces the \(x\)- and \(y\)-axes.

We will prove a theorem about the centralizer in \(\mathrm{GL}(V)\) of an irreducible matrix group, and all we really need is Proposition 0.2.

Theorem 1.7. Suppose that \(G\leq \mathrm{GL}(V)\) is an irreducible matrix group over an algebraically closed field \(k\), and let \(x\) be an element of \(\mathrm{GL}(V)\). If \(x\) centralizes \(G\), then \(x\) is a scalar matrix.

Proof: Since \(x\) is a matrix, it has an eigenvalue \(\lambda\) by Proposition 0.2. By multiplying \(x\) by the scalar matrix \(\lambda^{-1}I\) (which doesn't affect whether \(x\) is a scalar matrix, or whether \(x\) centralizes \(G\)) we may assume that \(\lambda=1\). Let \(W\) denote the \(1\)-eigenspace of the action of \(x\) on \(V\). (That is, let \(W\) be the set of all eigenvectors for the eigenvalue \(1\).)

Let \(g\in G\) and \(w\in W\). Since \(x\) centralizes \(G\), \(xg=gx\). On the one hand, since \(wx=w\), we have that\[ w(xg)=wg.\]On the other, \(w(gx)=(wg)x\), and this must equal \(wg\). Therefore \(wg\) is fixed by \(x\), i.e., lies in \(W\). In particular, this means that \(wg\in W\), so \(W\) is \(g\)-invariant, hence \(G\)-invariant as \(g\) was chosen arbitrarily.

However, \(G\) is irreducible, so \(W=0\) (not possible as \(1\) is an eigenvalue of \(x\)) or \(W=V\), i.e., \(x=I\), as claimed. □

From this we can prove a few nice properties of irreducible matrix groups.

Corollary 1.8. Let \(k\) be an algebraically closed field, \(V\) be a \(k\)-vector space, and \(G\leq \mathrm{GL}(V)\) be an irreducible matrix group.

1. (i) \(Z(G)\) consists of scalar matrices.
2. (ii) If \(Z(G)\) is finite then it is cyclic.
3. (iii) If \(G\) is abelian then \(\dim(V)=1\).

Proof: For the first part, note that \(Z(G)\) centralizes \(G\), and then apply Theorem 1.7 to get that \(Z(G)\) consists of scalar matrices. For the second part, note that the scalar matrices are isomorphic to \(k^\times\), and as every finite subgroup of that is cyclic by Lemma 0.1, we are done. For the third part, \(G\) consists of scalar matrices itself as \(G=Z(G)\), so in particular every subspace is \(G\)-invariant, so that \(\dim(V)=1\). □

Example 1.9. In the square case from our previous example, the \(180^\circ\) rotation lies in the centre of \(D_8\), and indeed it is represented by the matrix\[\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.\]In a \(3\)-dimensional reducible matrix group \(D_8\) (with the square held in the \(x,y\)-plane) the rotation is given by the matrix\[\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix},\]which of course is not a scalar matrix, but then \(G\) is not irreducible.