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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.

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  • $\begingroup$ “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 $\mathbb{Z}_3:\mathbb{Z}_4$? $\endgroup$
    – Groups
    Jun 23, 2019 at 2:59
  • $\begingroup$ Yes, I edited the question to make clear that that is the only group that I need help on. Thanks for your comment. $\endgroup$
    – vginhi
    Jun 23, 2019 at 3:59
  • $\begingroup$ 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/… $\endgroup$
    – reuns
    Jun 23, 2019 at 5:43

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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: $$ d\mapsto \begin{bmatrix}{ \omega \ 0 \\ 0 \ \omega^2 }\end{bmatrix}, \ \ \ f\mapsto \begin{bmatrix}{ 0 \ -1 \\ 1 \ \ \ \ \ 0 }\end{bmatrix} $$

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