Characterizations of the Orthogonal Groups _ Tapp, Ch. 3, Section 2

[Math Amateur]

I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 2 in Chapter 3, namely:

"2. Several Characterizations of the Orthogonal Groups".

I need help in fully understanding some important remarks following Proposition 3.10.

Section 2 in Ch. 3, including Proposition 3.10 reads as follows:https://www.physicsforums.com/attachments/3996
https://www.physicsforums.com/attachments/3997

Near to the bottom of the above text, after the statement of Proposition 3.10, we read:" ... ... Since $$U(n)$$ is isomorphic to its image, $$\rho_n ( U(n) )$$, ... ... "My question is how do we know that $$U(n)$$ is isomorphic to its image, $$\rho_n ( U_n) )$$ ... ... indeed, further ... how do we rigorously prove that $$U(n)$$ is isomorphic to its image, $$\rho_n ( U(n) )$$?I note in passing that Proposition 2.2 on page 25 of Tapp's book (see below for details) proves that $$\rho_n$$ is a linear transformation ... and I also note that intuitively $$\rho_n$$ would seem to be injective (but how on Earth do we prove it?) ... but I have no idea of how to prove the surjectivity of $$\rho_n$$ in the case of $$U(n) $$... ... I hope someone can help ...

Peter
***NOTE***

Tapp introduces $$\rho_n$$ in Section 1 of Ch. 2 (pages 24-25) ... so I am providing these pages as follows:View attachment 3998
View attachment 3999

 

[Fallen Angel]

Hi Peter,

In my first replie in post
http://mathhelpboards.com/linear-abstract-algebra-14/rotations-complex-matrices-real-matrices-proof-tapp-proposition-2-2-a-14380.html

I told you it was easy to check that $\rho_{n}$ was injective but NOT SURJECTIVE, reflect on it taking $\rho_{1}$ as an example.

What you have in Proposition 2.2 is that $\rho_{n}:\mathcal{M}_{n}(\Bbb{C})\longrightarrow \mathcal{M}_{2n}\Bbb{R}$ is a ring homomorphism, so we can apply first isomorphis theorem, which says that
$\begin{array}{cccc}\bar{\rho_{n}}:&\mathcal{M}_{n}(\Bbb{C})/Ker(\rho_{n}) &\longrightarrow& Im(\rho_{n})\\
& A+Ker(\rho_{n}) & \mapsto & \rho_{n}(A) \end{array}$

Is a ring isomorphism, but we know that $\rho_{n}$ is injective, so $Ker(\rho_{n})=\{0\}$

 

[Math Amateur]

Hi Peter,

In my first replie in post
http://mathhelpboards.com/linear-abstract-algebra-14/rotations-complex-matrices-real-matrices-proof-tapp-proposition-2-2-a-14380.html

I told you it was easy to check that $\rho_{n}$ was injective but NOT SURJECTIVE, reflect on it taking $\rho_{1}$ as an example.

What you have in Proposition 2.2 is that $\rho_{n}:\mathcal{M}_{n}(\Bbb{C})\longrightarrow \mathcal{M}_{2n}\Bbb{R}$ is a ring homomorphism, so we can apply first isomorphis theorem, which says that
$\begin{array}{cccc}\bar{\rho_{n}}:&\mathcal{M}_{n}(\Bbb{C})/Ker(\rho_{n}) &\longrightarrow& Im(\rho_{n})\\
& A+Ker(\rho_{n}) & \mapsto & \rho_{n}(A) \end{array}$

Is a ring isomorphism, but we know that $\rho_{n}$ is injective, so $Ker(\rho_{n})=\{0\}$

Hi Fallen Angel,

Thanks for the post ...

MY APOLOGIES! I made what was essentially a typo in my post ... I meant "injective" ... but for some strange reason that I cannot explain (mind you it is late here in Tasmania ... ), I typed "surjective" ... ...

I have now edited my post ...

Again ... sorry about the error ... now reflecting on $$\rho_n$$ as a ring homomorphism ... ...

Thanks for your help ...

Peter

 

[Math Amateur]

Hi Fallen Angel,

Thanks for the post ...

MY APOLOGIES! I made what was essentially a typo in my post ... I meant "injective" ... but for some strange reason that I cannot explain (mind you it is late here in Tasmania ... ), I typed "surjective" ... ...

I have now edited my post ...

Again ... sorry about the error ... now reflecting on $$\rho_n$$ as a ring homomorphism ... ...

Thanks for your help ...

Peter

Thanks for the help in showing that $$\overline{ \rho}_n$$ is an isomorphism ... allows me to move on through Tapp's textbook with more confidence ...

Appreciate your help!

Peter***EDIT***

Alternatively ... Presumably we could have used the First Isomorphism Theorem for Groups ... ... is that correct?

Peter