Conjugacy classes, irreducible representations, character table of $D_{10}$ (order 20)

Question

$D_{10}=\langle r,s \rangle$ is the dihedral group of order 20 . I have been struggling a bit with this question, particularly c, regarding values in the character table

(a). Find the conjugacy classes of $D_{10}$

Attempt at (a): G={${1, r, ..., r^9, s, rs, ..., r^9s}$}, $r^ir^j(r^i)^{-1}=r^j$ and $(r^is)r^j(r^is)^{-1}=r^{-j}$ so this means that the conjugacy class of $r^j$ is {$r^j, r^{-j}$} . Also, $r^i(r^js)(r^i)^{-1}=r^{2i}(r^js)$ and $r^is(r^js)(r^is)^{-1}=r^{i}sr^jss^{-1}r^{-i}=r^{2i-j}s=r^{2(i-j)}(r^js)$ So, the conjugacy class of $r^js$ is {$r^{2i}(r^js) | i=0, ..., 9$}

(b). List possible dimensions of all irreducible representations of $D_{10}$ and find the number of irreducible representations of each dimension.

Attempt at (b): G is finite so there will be finitely many irreducible representations. The sum of squares of dimensions of representations is equal to $|G|=20$, and the dimensions divide $|G|=20$. Hence possible dimensions are: $1, 2, 4, 5, 10$. I am not sure of the number of irreducible representations of each.

(c). Give the values of one row of the character table of $D_{10}$ corresponding to a character of degree $2$.

Attempt at (c) : I need help to do this one.

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Answer 1

(a) You have $r^{10}=1$, $s^2=1$ (so $s=s^{-1}$), and $rsrs=1$ (so $rs=sr^{-1}$ and $sr=r^{-1}s$).

For conjugates of $r^k$: Notice that $r^jr^kr^{-j}=r^k$ and $(r^js)r^k(r^js)^{-1}=r^jsr^ksr^{-j}=r^{j-k}ssr^{-j}=r^{-k}$. So $r^k$ is conjugate to itself and $r^{-k}$. We get conjugacy classes: $\{1\}$, $\{r,r^9\}$, $\{r^2,r^8\}$, $\{r^3,r^7\}$, $\{r^4,r^6\}$, and $\{r^5\}$.

For conjugates of $r^ks$: First, $r^j(r^ks)r^{-j}=r^{j+k}sr^{-j}=r^{2j+k}s$. Also, $(r^js)(r^ks)(r^js)^{-1}=r^jsr^kssr^{-j}=r^jsr^{k-j}=r^{2j-k}s$. Thus the conjugating can change a power of $r$ by an even amount. We thus get 2 conjugacy classes: $\{s,r^2s,r^4s,r^6s,r^8s\}$ and $\{rs,r^3s,r^5s,r^7s,r^9s\}$.

So there are 8 classes in all. These include the 2 singleton conjugacy classes are $\{1\}$ and $\{r^5\}$ so that the center is $Z(D_{10})=\{1,r^5\}$. The class equation is $20=1+1+2+2+2+2+5+5$.

(b) The dimensions of the irreducibles do need to divide $20$ so $1,2,4,5,10,20$. However, as you mentioned, the sum of squares of dimensions must add to 20. This means $4,5,10,$ and $20$ are too big since we must have 8 irred. reps. For example: $4^2=16$ which would mean $1^2+1^2+1^2+1^2+4^2=20$ and we only have 5 irred. reps. Thus all dimensions are 1 and 2. The only way to have 8 of these square and add to 20 is: $1^2+1^2+1^2+1^2+2^2+2^2+2^2+2^2=20$ (four degree 1 reps and four degree 2 reps).

(c) Dihedral groups can be represented by rotation/reflection matrices. For example: $r \mapsto \begin{bmatrix} \cos(2\pi/10) & -\sin(2\pi/10) \\ \sin(2\pi/10) & \cos(2\pi/10) \end{bmatrix}$ and $s \mapsto \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ (representing $s$ as a reflection across the $x$-axis). To see details about the characters of such a representation check out this page (look for "The linear representation theory of dihedral groups of even degree"). Once you have one of these representations in hand, you can manually compute the character.

However, usually there are quicker ways to compute the entries in a character table (like using the orthogonality relations). It's difficult to advise you since I'm not sure what tools you've learned so far.

I hope this helps!