These groups are not isomorphic but their characters are equal?

Question

The exercise is to compute the table of characters of two groups, D4 (group of symmetries of a square) and H4 (quaternions $\begin{pmatrix} c && -\overline{d} \\ d && \overline{c}\end{pmatrix}$ with $c=0$ or $d=0$ and $c, d \in \{0, 1, -1, i, -i\} $ )

These two groups are non commutative, and of size 8. I found that calling one element of order 4 "a", we can write the elements of the groups $\{1, a, a^2, a^3, b, ab, a^2b, a^3b\}$ with $a^4 = 1$ and $ba=a^3b$.

In D4, $b^2=1$ but in H4, $b^2=a^2$. Because of this (number of elements of order 2 or 4) I state that D4 and H4 are not isomorphic.

The conjugaison classes are: $\{1\}, \{a, a^3\}, \{a^2\}, \{b, a^2b\}, \{ab, a^3b\}$

Looking for characters:

4 characters have to be of dimension 1, and 1 of dimension 2.

In dimension 1, $\chi$ has to be commutative because $\mathbb{C^*}$ is, so setting $\chi(a)$ and $\chi(b)$ forces the rest. As $\chi(a)^2=1$ and other things, I don't have much choice, this gives me:

• $\chi_0(a)=1$ and $\chi_0(b)=1$ ($\chi_0=1$)
• $\chi_1(a)=1$ and $\chi_1(b)=-1$
• $\chi_2(a)=-1$ and $\chi_2(b)=-1$
• $\chi_3(a)=-1$ and $\chi_3(b)=1$

In dimension 2, let $\rho_a = \begin{pmatrix} i && 0 \\ 0 && -i\end{pmatrix}$ and $\rho_b = \begin{pmatrix} 0 && 1 \\ 1 && 0\end{pmatrix}$ be the representation operators. This gives

• $\chi_4(1)=2$, $\chi_4(a^2)=-2$ and $\chi_4=0$ otherwise

QUESTION

When I check the scalar product, it looks ok ($<\chi_i, \chi_j> = 0$ if $i \neq j$) so I have the impression that I found the 5 representations and that Im correct. But D4 and H4 are non isomorphic and their characters are equal! Is it normal? If not where am I wrong?

Answer 1

You' re right that $D_8$ (dihedral group of order 8) and $Q_{8}$(the group of quaternions) have the same character table still they are non-isomorphic groups. This example can be generalized to a more large class of group where this phenomenon happen. These are the groups of order $p^{3}$ for any odd-prime p. You can use semi-direct product to see that there are exactly 2 non-abelian non-isomorphic groups. Then using the techniques of chracter theory you can show that thry have the same character table.