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As we all know that the irreducible representation for Heisenberg group can be classified easily when the group is over a finite field $\mathbb{F}_q$, where $q=p^n$ and $p$ is a prime greater than $2$. The Heisenberg group is defined to be $$\left\{\left.\begin{pmatrix}1 & \mathbf{x}& t\\0 & I_n &\mathbf{y} \\ 0 & 0 & 1\end{pmatrix}\right|t\in\mathbb{F}_q,\mathbf{x},\mathbf{y}\in\mathbb{F}_q^n \right\}$$ with matrix multiplication. The way to classify them is to use the isomorphism $$\varphi:\begin{pmatrix}1 & \mathbf{x}& t\\0 & I_n &\mathbf{y} \\ 0 & 0 & 1\end{pmatrix} \to (t-\frac{1}{2}\mathbf{x}\cdot \mathbf{y},\mathbf{x},\mathbf{y}^T).$$ But in characteristic $2$, the $1/2$ does not make any sense. There's a way to treat the case here, but it seems hard to classify irreducible representation for the group. So my question is:

Are there any references for the classification of irreducible representation of characteristic $2$? Or can you give me some hints of how to do so?

Thanks for your help!

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Heisenberg groups over a finite field $\mathbb{F_q}$ with $q=2^m$ are abelian and its representations are all one-dimentional, i.e., characters. The classification of irreducible representations is given (among other references) in section $3$ of the article The Representations of the Heisenberg Group over a Finite Field by M. Misaghian. See section $2$ for the definition of the Heisenberg groups, also in characteristic $2$.

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    $\begingroup$ Are there different definitions of the Heisenberg group around? The group defined above is not abelian when $q$ is even and $n \ge 1$, and this agrees with the definition on the Wikipedia page, which says that the group is $D_8$ when $n=1$ and $q=2$. $\endgroup$
    – Derek Holt
    Commented Aug 26, 2014 at 15:16
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    $\begingroup$ The author of the article define the Heisenberg group to be $(x,a)(x',a')=(x+x',a+a'+\langle x,x'\rangle)$, which differs my definition when $p=2$. But anyway, this definition is still valid and gives a different group. $\endgroup$ Commented Aug 28, 2014 at 4:30

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