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Let $$D_6=\langle a,b| a^6=b^2=1, ab=ba^{-1}\rangle$$ $$D_6=\{1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}$$

I would like to compute its character table and its irreducible representations.

I will explain what I have done so far and I will add some doubts I had while doing this.

MY ATTEMPT

  1. Compute conjugacy classes. $$C_1=\{e\}, C_2=\{a,a^5\},C_3=\{a^2,a^4\}$$$$C_4=\{a^3\},C_5=\{b,a^2b,a^4b\},C_6=\{ab,a^3b,a^5b\}$$

  2. Find $1$-dimensional representations. Since $D_6/\{a,a^5\}\cong \mathbb{Z}_2$, we have one more representation apart from $\alpha_1=id$. That is $$\alpha_2: G \longrightarrow \mathbb{C}: a \mapsto 1, b\mapsto -1$$ Again using irreducible representations from quotient group by normal subgroup, I considered $G/\{\overline{1},\overline{a},\overline{b},\overline{ab}\}\cong \mathbb{Z}_2\times\mathbb{Z}_2$ (since it is abelian). Then from here I obtained $$\alpha_3:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto 1$$ $$\alpha_4:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto -1$$

  3. Find $2$-dimensional representations. I have seen in my notes that for $D_n$ we can define $2$-dimensional representations: $$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\ sin(\frac{2\pi}{n}) &cos(\frac{2\pi}{n})\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$ Hence my $$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$

  4. Build my character table. \begin{array}{|c|c|c|c|} \hline & C_1 & C_2 &C_3 &C_4 &C_5 &C_6 \\ \hline \chi_1& 1 & 1 &1 &1 &1&1 \\ \hline \chi_2& 1 & 1 &1 &1 &-1 &-1 \\ \hline \chi_3& 1 & -1 &1 &-1 &1 &-1 \\ \hline \chi_4& 1 & -1 &1 &-1 &-1 &1 \\ \hline \chi_5& 2 & 1 &-1 &-2 &0 &0 \\ \hline \chi_6& 2 & -1 &-1 &2 &0 &0 \\ \hline \end{array}

where I have computed $\chi_6$ by the orthogonality formula $(\chi_6|\chi_j)=\delta_{6,j}$.

QUESTIONS

  1. Is $D_6/\{a^2,a^4\}$ really abelian? I can not see it clearly.
  2. My first question comes when I have to find $2$-dimensional irreducible representations. I have find them because I have seen it in my notes. But how could I get $\alpha_5$ and $\alpha_6$ without knowing the special case of $D_n$. I know that I also could get it from $S_3$ (one of them). But I have again the same problem, if you are looking for $2$-dimensional irreducible representations of $S_3$, how do you find them? (Both).

  3. Now consider $X$ to be the set of the vertices of a regular $6$-gon and consider the action of $D_6$ on the set $X$ by restricting the usual action of $D_6$ on the $6$-gon to the set of vertices $X$. Let $\phi$ be the induced permutation representation (over $\mathbb{C}$) of $D_6$. I would like to write it as a sum of irreducible representations by computing the in-product of the irreducible characters with $\chi_{\phi}$. What should I do? I do not understand this induced permutation representation. Any help?

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    $\begingroup$ Consider the Klein four-group $K$ with generators $a'$ and $b'$. There is a homomorphism $\phi:D_n\to K$ sending $a$ and $b$ to $a'$ and $b'$. It is surjective with kernel generated by $a^2$. $\endgroup$ – Lord Shark the Unknown Jan 19 at 16:09
  • $\begingroup$ But is it injective? $\endgroup$ – idriskameni Jan 19 at 16:11
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    $\begingroup$ No, as I said, it is surjective, and as $D_6$ has more elements than $K$, it could hardly be injective. $\endgroup$ – Lord Shark the Unknown Jan 19 at 16:12
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In general, $D_n$ is a group of order $2n$ with a cyclic subgroup $C_n$ of order $n$ generated by $a$ say. Also $D_n$ contains $b$ with $b^2=1$ and $bab^{-1}=a^{-1}$.

One can obtain degree $2$ characters of $D_n$ by inducing from degree $1$ characters of $C_n$. For each $j$ there is a representation $\rho_j$ of $C_n$ taking $a^k$ to $\zeta^{jk}$ where $\zeta=\exp(2\pi i/n)$. This induces to a degree $2$ character of $D_n$ via $$\chi_j(g)=\rho_j(g)+\rho_j(bgb^{-1})$$ where we set $\rho_j(g)=0$ for $g$ outside $C_n$. Then $$\chi_j(a^k)=\zeta^{jk}+\zeta^{-jk}=2\cos\frac{2\pi jk}n$$ and $$\chi_j(a^kb)=0.$$

This character $\chi_j$ is irreducible unless $\zeta^j=\pm1$. Together with the degree $1$ characters of $D_n$, the irreducible $\chi_j$ exhaust the characters of $D_n$.

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  1. It has four elements, so yes.

  2. Get the characters by messing around with orthogonality, then come up with a representation that does that.

  3. Label the vertices of your hexagon $a_1$ through $a_6$. Then take the vector space $V$ over $\mathbb{C}$ to be the set of formal $\mathbb{C}$-weighted sums of $a_1$ through $a_6$. Then we construct a representation of $D_6$ by an action on this vector space given by defining, for $\sigma \in D_6$, the action of $\sigma$ to be that which sends each $a_i$ to the $a_j$ that $\sigma$ sends $a_i$ to in the action on the hexagon. Calculate the character of this representation, then proceed as the question tells you to.

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  • $\begingroup$ What is the vector space $V$ to be the set of formal $\mathbb{C}$-weighted sums? $\endgroup$ – idriskameni Jan 19 at 16:14
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    $\begingroup$ @idriskameni Just that: take the elements of $V$ to be all elements of the form $\alpha_1 a_1 + \alpha_2 a_2 + \alpha_3 a_3 + \alpha_4 a_4 + \alpha_5 a_5 + \alpha_6 a_6$, with $\alpha_i \in \mathbb{C}$, and the obvious addition and scalar multiplication. $\endgroup$ – user3482749 Jan 19 at 16:15
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  1. Is $D_6/\{a^2,a^4\}$ really abelian? I can not see it clearly.

In your notation it looks like $\{a^2,a^4\}$ is denoting a conjugacy class and not a (normal) subgroup, so presumably you mean $D_6/\{1,a^2,a^4\}$. But yes its abelian.

  1. My first question comes when I have to find $2$-dimensional irreducible representations. I have find them because I have seen it in my notes. But how could I get $\alpha_5$ and $\alpha_6$ without knowing the special case of $D_n$. I know that I also could get it from $S_3$ (one of them). But I have again the same problem, if you are looking for $2$-dimensional irreducible representations of $S_3$, how do you find them? (Both).

To find irreducible representations of $S_3$ note there is a natural 3-dimensional representation as permutation matrices (the standard representation). There is a 1 dimensional sub-representation of this which is the span of the vector $(1,1,1)$. Split off this summand to get an 2d irrep.

  1. Now consider $X$ to be the set of the vertices of a regular $6$-gon and consider the action of $D_6$ on the set $X$ by restricting the usual action of $D_6$ on the $6$-gon to the set of vertices $X$. Let $\phi$ be the induced permutation representation (over $\mathbb{C}$) of $D_6$. I would like to write it as a sum of irreducible representations by computing the in-product of the irreducible characters with $\chi_{\phi}$. What should I do? I do not understand this induced permutation representation. Any help?

Remember $\chi_\phi(g) = tr(\phi(g))$. Since this character is coming from a group action there is also the interpretation as the number of fixed points, $\chi_\phi(g) = |\#\{x : g.x = x\}|$. For example if $g$ is a rotation then $\phi(g)$ has no fixed points so $\chi = 0$. Equivalently, $\phi(g)$ has 0s on the diagonal so $\chi_\phi(g) = tr(\phi(g)) = 0$. If $\phi(g)$ is a reflection then there will be fixed points and the character will be non-zero.

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  • $\begingroup$ Do you mean $D_6/\{1,a^2,a^4\}$? $\endgroup$ – idriskameni Jan 19 at 16:25
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    $\begingroup$ @idriskameni yes, fixed. $\endgroup$ – Ben Jan 20 at 9:21

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