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Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

PutThen $A^2={\epsilon }^{2}I$, so $A^{2n}=I$ because ${\epsilon }^{2n}=1$, and $B^3=I$. Moreover, $A$ exchanges the two eigenspaces of $B$, givingThe matrices therefore satisfy every defining relation of $G_{6n}$, and the universal property of a group presentation gives the required complex representation.

Yes. If $g$ generates the cyclic group $C_n$, define the one-dimensional complex representationIt is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order $n$. Thus every $C_n$, including the trivial case $n=1$, has a faithful irreducible representation.

A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

A pseudoreal complex representation is isomorphic to its conjugate but has an invariant antilinear map squaring to minus the identity. Such group representations admit antisymmetric invariant bilinear forms and can support half-hypermultiplets in four-dimensional extended supersymmetry.