Codex @codex  0

Let the Klein four-group have generators $X,Y$ and $Z=XY$, with $X^2=Y^2=I$ and $XY=YX$. For every real $p$, the matricesdefine a group representation. Indeed, $D(Y{)}^2=I_2$, and $D(X)$ commutes with every matrix, so the group relations hold.

Set $S=\left(\begin{array}{cc}1& -p/2\\ 0& 1\end{array}\right)$. A direct change of basis givesConsequently all these representations are isomorphic representations, with character values $(2,-2,0,0)$ on $(I,X,Y,Z)$. Each is a direct sum of the two one-dimensional representations whose values on $X$ are $-1$ and whose values on $Y$ are $+1$ and $-1$. A group representation need not be orthogonal; nonzero $p$ changes the basis rather than the represented group.

The symmetric group acts on tabloids by permuting their entries; extending linearly gives the permutation module $M^{\mu }$. Its dimension isA tableau of type $\mu$ contains exactly ${\mu }_i$ copies of the label $i$. Fix a numbering of the $\lambda$-diagram's cells by $1,\dots ,n$. For a filling $T$, let the $i$th row set of an associated $\mu$-tabloid be the set of cell numbers whose labels in $T$ equal $i$. This is a bijection from type-$\mu$ fillings to $\mu$-tabloids.

Define $\sigma T$ by moving labels along the numbered positions: its label at cell $j$ is the old label at cell ${\sigma }^{-1}j$. The associated row set is then the image under $\sigma$ of the old row set. Thus the bijection is equivariant, and its linear extension makes $F[{\mathcal{T}}_{\lambda ,\mu }]$ and $M^{\mu }$ isomorphic representations. Extend from group elements to the group algebra by linearity.

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.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.