-----
> [!proposition] Proposition. ([[character tables of small groups]])
> ## $C_{2}=\{ e,x \}$
> Recall that, in an [[abelian group]], every element exists in a [[conjugate|conjugacy class]] of its own. So the [[character table]] for $C_{2}$ will have two columns, corresponding to $[e]$ and $[x]$. When searching for [[irreducible group representation]]s we begin by looking for $1$-dimensional [[group representation|representations]], i.e., [[group homomorphism|homomorphisms]] from $C_{2} \to \mathbb{C}^{\times}$. The trivial homomorphism that sends $x \mapsto 1$ yields the [[trivial group representation]]. The only other [[group homomorphism]] sends $x \to -1$ (because $\im x$ must have order $2$).
>
>
>
> | $C_{2}$ | $[e]$ | $[x]$ |
> |----|-----|-----|
> | $\chi_{\mathbb{1}}$ | 1 | 1 |
> | $\chi$ | 1 | -1 |
>
>
> ## $C_{3} = \{ e, x,x^{2} \}$
> $C_{3}$ is abelian, hence table has columns for $[e], [x], [x^{2}]$. The image of $x$ under a nontrivial [[group homomorphism|homomorphism]] $C_{3} \xrightarrow \phi \mathbb{C}^{\times}$ must have order $3$, i.e., $\phi^{3}(x)=1$ which happens iff $\phi(x)$ is a third [[root of unity]]: $\phi(x) \in \{ 1, \omega, \omega^{2} \}$. Then $\phi(x^{2})=\phi^{2}(x) \in \{ 1, \omega^{2}, \omega \}$.
>
>
>
> | $C_{3}$ | $[e]$ | $[x]$ | $[x^2]$ |
> |-----|-----|-------|-------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 |
> | $\chi$ | 1 | $\omega$ | $\omega^{2}$ |
> | $\chi^{2}$ | 1 | $\omega^{2}$ | $\omega$ |
>
> **Remark.** The use of $\chi^{2}$ is not just notation: in this special case, the function $\chi^{2}$ indeed takes on values that are squares of those of $\chi$. (In some sense, characters and 1D representations are the same thing.)
>
> **Remark.** We see that the [[standard representation of the dihedral group|standard representation]] of $C_{3}$ (as rotations) is not an [[irreducible group representation]] because its [[character of a representation|character]], $\chi_{\text{std}}$, arises as the sum of $\chi$ and $\chi^{2}$. (See Notability)
>
> ## $C_{4} = \{ e, x,x^{2}, x^{3}\}$
> $C_{4}$ is [[abelian group|abelian]], so our table has columns for $[e], [x], [x^{2}], [x^{3}]$. The image of $x$ under a [[group homomorphism|homomorphism]] $C_{4} \xrightarrow \phi \mathbb{C}^{\times}$ must have [[order of an element in a group|order]] $1$ (trivial case), order $2$ or order $4$. Such elements in $\mathbb{C}$ are the fourth [[roots of unity]] $\{ 1,-1,i,-i \}$. Sending $x$ to each of these elements gives four different homomorphisms from $C_{4}$ to $\mathbb{C}^{\times}$, hence four [[irreducible group representation]]s.
>
>
>
> | $C_{4}$ | $[e]$ | $[x]$ | $[x^2]$ | $[x^3]$ |
> |-----|-----|-------|-------|-------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 |
> | $\chi$ | 1 | $-i$ | $-1$ | $i$ |
> | $\chi^{2}$ | 1 | $-1$ | $1$ | $-1$ |
> | $\chi^{3}$ | 1 | $i$ | $-1$ | $i$ |
>
>
> ## $C_{2} \times C_{2} = \{ e,a,b,ab \}$
> The [[Klein 4-group]] is [[abelian group|abelian]], so our table has columns for $[e], [a], [b], [ab]$. We need [[complex numbers|complex numbers]] satisfying the relations $x^{2}=e, y^{2}=e, xy=yx$. In addition to the trivial homomorphism, we could map $a \mapsto 1$, $b \mapsto -1$. We could also do $a \mapsto -1$, $b \mapsto 1$. Or we could do $a \mapsto -1, b \mapsto -1$. These give the [[character table]]
>
> [!proof]- Proof. ([[character tables of small groups]])
> | $C_{2} \times C_{2}$ | $[e]$ | $[a]$ | $[b]$ | $[ab]$ |
> |-----|-----|------|------|------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 |
> | $\chi_{1}$ | 1 | $1$ | $-1$ | $-1$ |
> | $\chi_{2}$ | 1 | $-1$ | 1 | $-1$ |
> | $\chi_{3}$ | 1 | $-1$ | $-1$ | 1 |
>
>
>
>
> ## $S_{3}$
> The [[conjugate|conjugacy classes]] of $S_{3}$ are $\{ e \}, \{ x, x^{2} \}=[x], \{ y, xy, x^{2}y \}=[y]$. [[character of a representation#^9ecc4a|We did this here]]:
>
>
>
> | | $1$ | $[x]$ | $[y]$|
> |----------|-----|----------|---------------|
> | $\chi _T$ | $1$ | $1$ | $1$ |
> | $\chi _\Sigma$ | $1$ | $1$ | $-1$ |
> | $\chi _A$ | $2$ | $-1$ | $0$ |
>
>
>
> ## $D_{4}$
> The [[conjugacy classes of the dihedral group|conjugacy classes of]] $D_{4}$ are $\{ e \}=[e], \{ x^{2} \}=[x^{2}], \{ x,x^{3} \}=[x], \{ y, yx^{2} \}=[y], \{ yx,yx^{3} \}=[yx]$.
>
> We begin by looking for 1-dimensional representations since clearly these are [[irreducible group representation|irreducible]]. So, we want to find [[group homomorphism|homomorphisms]] $D_{4} \xrightarrow{\phi} \mathbb{C}^{\times}$ that satisfy the [[dihedral group]] relations.
> ![[CleanShot 2023-11-27 at 19.35.58 1.jpg]]
> Thus we have four 1-dimensional irreps $\rho_{\mathbb{1}}, \rho_{2}, \rho_{3}, \rho_{4}$ of $D_{4}$. We'll denote the nontrivial corresponding characters $\chi_{\rho}=\chi_{(\rho(x), \rho(y))}$ since each $\rho$ is characterized by where it sends $x$ and $y$. This yields the partial [[character table]]
>
>
>
> | $D_{4}$ | $[e]$ | $[x^2]$ | $[x]$ | $[y]$ | $[yx]$ |
> |----------------------------|--------|---------|-------|-------|--------|
> | $\chi_\mathbb{1}$ | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
> | $\chi_5$ | 2 | | | | |
>
> where we are using the fact that [[number of irreps equals number of conjugacy classes]] to conclude that there is just one irreducible representation $\rho_{5}$ left to analyze with $\dim \rho_{5}=2$ (since $4 + \dim ^{2} \rho_{5}=|D_{4}|$). We know that, viewed as a vector, $\chi_{5}$ needs to be [[orthogonal]] to each other character. In order for columns 1 and 2 to be orthogonal we need to have $4 + 2 \chi_{5}(x^{2})=0$ hence $\chi_{5}(x^{2})=-2$.
>
>
>
> | $D_{4}$ | $[e]$ | $[x^2]$ | $[x]$ | $[y]$ | $[yx]$ |
> |----------------------------|--------|---------|-------|-------|--------|
> | $\chi_\mathbb{1}$ | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
> | $\chi_5$ | 2 | -2 | | | |
>
> Then we need to have $0 + -2 \chi_{5}(x)=0$ so $\chi_{5}(x)=0$ and similarly reasoning yields $\chi_{5}(y)=0$ and $\chi_{5}(yx)=0$. Thus $\chi_{5}$ corresponds to the character for the [[standard representation of the dihedral group|standard representation]] of $D_{4}$:
>
>
>
> | $D_{4}$ | $[e]$ | $[x^2]$ | $[x]$ | $[y]$ | $[yx]$ |
> |----------------------------|--------|---------|-------|-------|--------|
> | $\chi_\mathbb{1}$ | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
> | $\chi_\text{std}$ | 2 | -2 | 0 | 0 | 0 |
>
> ## $Q_{8}$
> The [[conjugacy classes of the quaternion group]] are $\{ \v 1 \}=[\v 1], \{ - \v 1 \}=[-\v 1], \{ \v i, - \v i \} = [\v i], \{ \v j, - \v j \}=[\v j], \{ \v k, -\v k \}=\v k.$ We begin by looking for 1-dimensional representations since clearly these are [[irreducible group representation|irreducible]]. So, we want to find [[group homomorphism|homomorphisms]] $Q \xrightarrow{\phi} \mathbb{C}^{\times}$ that satisfy the [[quaternion group]] relations $\v i ^{4}=e, \v i^{2}=\v j ^{2}, \v i \v j \v i ^{-1} \v j=e$.
>
> ![[CleanShot 2023-11-27 at
[email protected]|500]]
> ![[CleanShot 2023-11-27 at 20.53.07.jpg|500]]
>
> Thus we have four 1-dimensional irreps $\rho_{\mathbb{1}}, \rho_{2}, \rho_{3}, \rho_{4}$ of $Q_{8}$. We'll denote the nontrivial corresponding characters $\chi_{\rho}=\chi_{(\rho(\v i), \rho(\v j))}$ since each $\rho$ is characterized by where it sends $\v i$ and $\v j$. This yields the partial [[character table]]
>
>
>
> | $Q_{8}$ | $[\mathbf{1}]$ | $[-\mathbf{1}]$ | $[\mathbf{i}]$ | $[\mathbf{j}]$ | $[\mathbf{k}]$ |
> |----------------------------|----------------|-----------------|----------------|----------------|----------------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
> | $\chi_5$ | 2 | | | | |
>
> where we used the fact that $- \v 1 = \v i ^{2}, \v k= \v i \v j$, and $\chi_{5}$ must correspond to a representation with dimension 2. Now exploiting column orthogonality we find that the only option for filling in the rest of the table is
>
>
>
> | $Q_{8}$ | $[\mathbf{1}]$ | $[-\mathbf{1}]$ | $[\mathbf{i}]$ | $[\mathbf{j}]$ | $[\mathbf{k}]$ |
> |----------------------------|----------------|-----------------|----------------|----------------|----------------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
> | $\chi_5$ | 2 | -2 | 0 | 0 | 0 |
>
> This matches the [[character table]] of $D_{4}$!
>
>
> # $D_{6}$
> The [[conjugate|conjugacy classes]] of $D_{6}$ are $\{ e \}=[e]$, $\{ x^{3} \}=[x^{3}]$, $\{ x, x^{5} \}=[x], \{ x^{2}, x^{4} \}=[x^{2}]$, $\{ y, yx^{2}, yx^{4} \}=[y]$, $\{ yx,yx^{3}, yx^{5} \}=[yx]$. [[number of irreps equals number of conjugacy classes|So there are]] 6 irrep characters to analyze. From [[classification of irreps of the dihedral group over C]] we actually know the irreps of $D_{6}$, they are
> - $\rho_{\mathbb{1}}$
> - $\rho_{(1,-1)}$
> - $\rho_{(-1,1)}$
> - $\rho_{(-1,-1)}$
> - $\rho^{(1)}$
> - $\rho^{(1)}$
>
> where $\rho^{(1)}$ maps $\begin{align}
> x & \xmapsto{\rho^{(1)}} \begin{bmatrix}
> \cos \frac{2\pi}{6} & -\sin \frac{2\pi}{6} \\
> \sin \frac{2\pi}{6} & \cos \frac{2\pi}{6}
> \end{bmatrix} \\
> y & \xmapsto{\rho^{(1)}} \begin{bmatrix}
> 1 & 0 \\ 0 & -1
> \end{bmatrix}
> \end{align}$
> and $\rho^{(2)}$ maps $\begin{align}
> x & \xmapsto{\rho^{(2)}} \begin{bmatrix}
> \cos \frac{4\pi}{6} & -\sin \frac{4\pi}{6} \\
> \sin \frac{4\pi}{6} & \cos \frac{4\pi}{6}
> \end{bmatrix} \\
> y & \xmapsto{\rho^{(2)}} \begin{bmatrix}
> 1 & 0 \\ 0 & -1
> \end{bmatrix}.
> \end{align}$
> The character table follows by manually computing the characters...
>
>
>
> | $D_{6}$ | $[e] (1)$ | $[x^3] (1)$ | $[x] (2)$ | $[x^2] (2)$ | $[y] (3)$ | $[yx] (3)$ |
> |------------------------|-------|---------|-------|---------|-------|--------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 | 1 | 1 |
> | $\chi_{(1,-1)}$ | 1 | 1 | 1 | 1 | -1 | -1 |
> | $\chi_{(-1,1)}$ | 1 | -1 | -1 | 1 | 1 | -1 |
> | $\chi_{(-1,-1)}$ | 1 | -1 | -1 | 1 | 1 | -1 |
> | $\chi^{(1)}$ | 2 | -2 | 1 | -1 | 0 | 0 |
> | $\chi^{(2)}$ | 2 | 2 | -1 | -1 | 0 | 0 |
>
> # $C_{3} \rtimes C_{4}$
> $C_{3}$ is a [[normal subgroup]] of this [[group]]; we showed above it has character table
>
>
>
> | $C_{3}$ | $[e]$ | $[x]$ | $[x^2]$ |
> |-----|-----|-------|-------|
> | $\chi_{\mathbb{1}}$ | 1 | 1 | 1 |
> | $\chi$ | 1 | $\omega$ | $\omega^{2}$ |
> | $\chi^{2}$ | 1 | $\omega^{2}$ | $\omega$ |
>
>
> # $S_{4}$
> Let $K:=C_{2} \times C_{2}$. We we claim $S_{4} \cong K \rtimes S_{3}$. We already have seen the the disjoint products of two-cycles in $S_{4}$ form a [[normal subgroup]] isomorphic to $K$. And clearly $S_{3}$ is isomorphic to a [[subgroup]] of $S_{4}$ (e.g. the [[subgroup]] of permutations that fix the 4th letter). These [[subgroup]]s intersect trivially and so by [[cardinality of frobenius product of subgroups]] and [[internal semi-direct product]] we conclude $S_{4}=KS_{3}$ as required.
>
> Recall that in $S_{n}$, [[conjugation in the symmetric group is relabeling|permutations are conjugate iff they have the same cycle type]]. This means that the [[conjugate|conjugacy classes]] of $S_{4}$ are $[(1)], [(12)] , [(12)(34)], [(123)], [(1234)]$. Now the [[irreducible group representation]]s of $S_{3}$ [[lifting representations|irreps]] to [[irreducible group representation]]s of $S_{4}$ and we can record this in the [[character table]]. Below is the [[character table]] of $S_{3}$:
>
>
>
> | | $1$ | $[x]$ | $[y]$|
> |----------|-----|----------|---------------|
> | $\chi _\mathbb{1}$ | $1$ | $1$ | $1$ |
> | $\chi _\Sigma$ | $1$ | $1$ | $-1$ |
> | $\chi _A$ | $2$ | $-1$ | $0$ |
>
> Given an irrep $\tilde{\rho}$ with character $\tilde{\chi}$ of $S_{3} \cong S_{4} / K$ that we lift to an irrep $\rho$ with character $\chi$ of $S_{4}$ by defining $\rho(g)= \tilde{\rho}(gK)$, what does $\rho$ look like? [[lifting representations|What we know is that]] given $g \in S_{4}$, $\chi(g)=\tilde{\chi}(gK)$. The identity in the [[quotient group|quotient]] lifts to $(1)$ and $(12)(34)$, so these are all 1 in the table.
>
> $(12)$ goes with the transposition $y$
> $(123)$ to $x$
> $(1234)$ needs to go to a 1-cycle or 2-cycle, but it isn't in $K$ so must go to 2-cycle, so it too belongs with $y$. This gives the first 3 rows.
>
> Next, we think about the [[standard representation of the symmetric group]]. In particular, we know that the action of $S_{4}$ on $\mathbb{R}^{4}$ by [[permutation matrix|permutation matrices]] has an [[group-invariant subspace]] $\span(\b 1_{4})$ with [[group-invariant subspace admits group-invariant complement over C|invariant complement]] $\span^{\perp}(\b 1_{4})$ yielding the decomposition of $\mathbb{R}^{4}$ into [[irreducible group representation|irreps]] $\rho^{\text{perm}}=\rho^{(\b 1_{4})} \oplus \rho^{\text{std}}$ and in fact the action operates on that subspace as the identity. Hence in the [[block matrix]] corresponding to $\rho^{\text{perm}}_{\sigma}$, $\sigma \in S_{4}$, the singleton in the left block diagonal equals $1$. It follows that since $\chi^{\text{perm}}=\chi^{\b 1_{4}}+ \chi^{\text{std}}$ and that $\rho^{(\b 1_{4})}=\rho_{\mathbb{1}}$. Thus we have $\chi_{\text{std}}=\chi^{\text{perm}}-\chi_{\mathbb{1}}$. Recalling that the [[trace of a matrix|trace]] of a [[permutation matrix]] is just the number of fixed points of the corresponding [[permutation]], we can now compute the fourth row. The final row the fills itself using [[class function|orthogonality relations]].
>
>
>
> | $S_{4}$ | $[(1)]$ | $[(12)]$ | $[(12)(34)]$ | $[(123)]$ | $[(1234)]$ |
> |---------------------|---------|----------|--------------|-----------|------------|
> | $\chi_{\mathbb{\text{lift }1}}$ | 1 | 1 | 1 | 1 | 1 |
> | ${\chi}_{\text{lift }\Sigma}$ | 1 | -1 | 1 | 1 | -1 |
> | $\chi_{\text{lift }A}$ | 2 | 0 | 2 | -1 | 0 |
> | $\chi_\text{std}=\chi^{\text{perm}}-\chi_{\mathbb{1}}$ | 3 | 1 | -1 | 0 | -1 |
> | $\chi_\text{name5}$ | 3 | $-1$ | $-1$ | 0 | 1 |
>
> Let's now explore how each of these irreps' restriction to $A_{4}$ decomposes as a sum of representations of $A_{4}$. We discard the conjugacy classes $[(12)]$ and $[(1234)]$, and split the conjugacy class $[(123)]$ into two conjugacy classes $[(123)], [(123)^{2}]$. Following this, we see that the rows for $\chi_{\text{lift }\mathbb{1}}$ and $\chi_{\text{lift }\Sigma}$ are identical, as are the rows of $\chi_{\text{std}}$ and $\chi_{\text{name5}}$. So we are left with 3 representations to consider:
> - $\chi_{\mathbb{1}}$
> - $\chi_{\text{lift }A}$
> - $\chi_{\text{std}}$.
>
> We know that $A_{4}$ has $4$ conjugacy classes, so one of these must split into two irreps. n the [[character table]] of $A_{4}$ we already see $\chi_{\mathbb{1}}$ and $\chi_{\text{std}}$, so it isn't those. Hence $\chi_{\text{lift }A}$ splits into $\chi_{\text{lift }A}=\chi_{2}+\chi_{3}$, where
> $\chi_{2}=(1, 1, \omega, \omega^{2})$ and
> $\chi_{3}=(1,1,\omega^{2}, \omega)$.
>
> The only verification need is that $1+1=2$ and $\omega+\omega^{2}=-1$ (which is true because the 3rd [[roots of unity]] $1,\omega,\omega^{2}$ all sum to zero).
>
>
-----
####
----
#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```