# Getting real group representations from characters for a very small group (Order 12)

I am trying to understand the relationship between real, complex, and quaternion representations, and their characters. The answers by Jack Schmidt and Geoff Robinson to kcrisman’s question give a complete answer for the characters, but I am having a hard time trying to get an actual matrix. The group of order $$12$$ that I am interested in is of course the semidirect product of $$\mathbb{Z}/3$$ and $$\mathbb{Z}/4$$. The abelian groups of order $$12$$ are not a problem to understand, and the dihedral group is the symmetries of the hexagon, so it is not hard to construct the rotation and reflection matrices. The character table shows two $$2$$ dimensional representations, one of which is real (by the Frobenius-Schur indicator) and the other comes from the quaternions, but I have no idea how to construct matrices. Help, or a reference, or another stack exchange would be appreciated.

• “The group of order 12 is of course the semidirect product of Z/3 and Z/4." There are $5$ groups (up to isomorphism) of order 12, not all of them are semidirect product of $Z/3$ and $Z/4$. Do you mean you only consider the group $\mathbb{Z}_3:\mathbb{Z}_4$? Jun 23, 2019 at 2:59
• Yes, I edited the question to make clear that that is the only group that I need help on. Thanks for your comment. Jun 23, 2019 at 3:59
• For the 2-dimensional representations of the dihedral group for each $a$ send $n \in Z/6$ to $\pmatrix{\zeta_6^{an} & 0 \\ 0 & \zeta_6^{-an}}$ and the reflection to $\pmatrix{0 & 1 \\ 1 & 0}$ groupprops.subwiki.org/wiki/… Jun 23, 2019 at 5:43

You are interested in the group $$\langle d,f \mid d^3=f^4=1, f^{-1}df=d^{-1}\rangle$$.
There are two $$2$$-dimensional characters. Both take value $$(-1)$$ on $$d, d^{-1}$$, and value $$0$$ on the conjugates of $$f$$ and $$f^{-1}$$.
One takes value $$2$$ on $$f^2$$, so that $$f^2$$ is in the kernel, and so we are just looking at the two dimensional irreducible representation of $$D_6=S_3$$. If you wish to see this in terms of real matrices then I think you'll find $$d\mapsto \begin{bmatrix}{ 0 \ -1 \\ 1 \ -1 }\end{bmatrix}, \ \ \ f\mapsto \begin{bmatrix}{ 0 \ 1 \\ 1 \ 0 }\end{bmatrix}$$ will work for you.
For the other representation, the one where the character of $$f^2$$ is $$(-2)$$, then as you say we can't achieve it over $$\mathbb{R}$$. However, these complex matrices (with $$\omega$$ a primitive cube root of unity) will give a representation: $$d\mapsto \begin{bmatrix}{ \omega \ 0 \\ 0 \ \omega^2 }\end{bmatrix}, \ \ \ f\mapsto \begin{bmatrix}{ 0 \ -1 \\ 1 \ \ \ \ \ 0 }\end{bmatrix}$$