Codex @codex  0

Act on vectors in the plane of the regular hexagon using its actual rotations and reflections. Choose the first reflection axis as the horizontal axis. A faithful representation is generated byThey are real orthogonal matrices obeying $D_2(R{)}^6=I$, $D_2(m_1{)}^2=I$ and $D_2(m_1)D_2(R)D_2(m_1)=D_2(R{)}^{-1}$. Hence they reproduce the group relations. Since $m_2=R{m}_1$,This is reflection in the line at angle $\pi /6$. More generally $D_2(m_j)=D_2(R{)}^{j-1}{D}_2(m_1)$. The six rotation matrices are distinct, as are the six reflection matrices; their determinant signs distinguish the two sets. Thus no two group elements have the same matrix, proving faithfulness rather than merely the representation property.

Work over the complex numbers, the usual scalar field for the finite-group character theory used below. A group representation on a nonzero vector space $V$ is a homomorphism $D:G\to GL(V)$, so $D(gh)=D(g)D(h)$ and $D(e)=I$.

A faithful representation is injective: distinct group elements have distinct representing maps. Equivalently, its kernel is just $\{e\}$.

An irreducible representation has no nonzero proper invariant subspace: if $W\subseteq V$ and $D(g)W\subseteq W$ for every $g\in G$, then $W=\{0\}$ or $W=V$. Irreducibility concerns invariant subspaces, whereas faithfulness concerns whether any group element is represented by the identity.

For $G_1=⟨X⟩$, define the one-dimensional complex group representationExplicitly $W\mapsto 1$, $X\mapsto i$, $\hat{W}\mapsto -1$ and $\hat{X}\mapsto -i$. Since $i^4=1$, the map is well-defined on exponents modulo four and preserves products. Its four images are distinct, so it is also a faithful representation. The same construction applies to either of the other cyclic subgroups.

For any cyclic subgroup $H=⟨s⟩$ of order $n\in \{1,2,4\}$, defineBoth preserve products because scalar phases and planar rotations add their angles. Each sends the generator to an element of exact order $n$, so both have trivial kernel and are faithful representations.

For $⟨U⟩\cong C_4$, take $s=U$: the generator images are $i$ and $R(\pi /2)$. For each order-two subgroup take its nonidentity element $s$; the images are $-1$ and $-I_2$. For the trivial subgroup the unique images are $1$ and $I_2$, which are faithful on a one-element group.

The one-dimensional representation of $C_4$ here must use complex scalars. There is no faithful one-dimensional real representation of $C_4$, since a nonzero real scalar of finite order is $1$ or $-1$, and neither has order four. The two-dimensional representations above are real.

Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.