# Can we build a D4 matrix representation with ${\mathbb Z_2}^{4\times 4}$ matrices?

If we look at the Dihedral 4 group. There exists a trivial matrix representation that also makes it very easy to define group action on vectors: simply use 2x2 rotation matrices for the cyclic part and a reflection matrix for the reflection part. Then group action will be given by simply multiplying representation matrix right onto coordinate vector.

However I would like to build a matrix representation using a simpler field. Say for example $$\mathbb Z_2$$.

Now what I know is that the "cyclic part" of this group can be done with permutation matrices, generator:

$$C_G = \left[\begin{array}{cccc}&1&&\\&&1&\\&&&1\\1&&&\end{array}\right]$$

The non-reflected rotations could then have matrices $$\{{C_G}^0, {C_G}^1,{C_G}^2,{C_G}^3\}$$

But how can we do the reflection part? And further: How can we afterwards define group action (working on $$\mathbb R^2$$) with this new matrix representation?

Generally, if $$G$$ embeds into $$S_n$$ then for any field $$F$$, $$G$$ embeds into $$GL_N(F)$$ as follows (note that I prefer right actions, but everything can be done on the left):

For convenience, identify $$G$$ with some isomorphic subgroup of $$S_n$$ so $$G\le S_n$$.

Fix a basis $$e_1,\ldots,e_n$$ of $$F^n$$. An embedding $$\psi:G\hookrightarrow GL_n(F)$$ is then given by $$e_i\psi(g)=e_{i^g}$$ (extended linearly).

In your case, a choice for one reflection would be

$$\left(\begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{matrix}\right)$$

The other reflections would be the conjugates of this by your rotations.

An action through linear maps on $$\mathbb{R}^2$$ is equivalent to an isomorphism with a subgroup of $$GL_2(\mathbb{R})$$ which you noted at the start of your question.