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This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $

In a paper of Wall (page 33), it is mentioned that the order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$.

How to prove this?

Any help will be appreciated.

This may be an easy exercise but I am not getting it. Let $F_q$ be a finite field with $q$ elements and $F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \{ A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \} $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{{n^2-n}/2} \Pi _{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

This may be an easy exercise but I am not able to get it. Let $F_q$ be a finite field with $q$ elements and $F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \{ A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \} $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{{n^2-n}/2} \Pi _{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

This may be an easy exercise but I am not getting it. Let $F_q$ be a finite field with $q$ elements and $F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \{ A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \} $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{{n^2-n}/2} \Pi _{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.