# irreducible representations and character table of $D_6$

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?

• 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 $a^2$. – Lord Shark the Unknown Jan 19 at 16:09
• But is it injective? – idriskameni Jan 19 at 16:11
• No, as I said, it is surjective, and as $D_6$ has more elements than $K$, it could hardly be injective. – Lord Shark the Unknown Jan 19 at 16:12

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 $$D_n$$.

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.

• What is the vector space $V$ to be the set of formal $\mathbb{C}$-weighted sums? – idriskameni Jan 19 at 16:14
• @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. – user3482749 Jan 19 at 16:15
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.

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