Notes to My Cat, equalizer in Grp and O(n,R)

 

This essay is aim at showing that where is $O(n,R)$$O(n,R)$ comes from and hence provided a natural proof for $O(n,R)$$O(n,R)$ is a subgroup of ${\mathrm{GL}}_n(R)$$\mathrm{GL}_n(R)$.

Let $R$$R$ be a commutative ring and $G{L}_n(R)=U\circ M_{n\times n}(R)$$GL_n(R)=U\circ M_{n\times n}(R)$, where $U$$U$ is the unit functor.

Let $G^{\mathrm{op}}$$G^{\mathrm{op}}$ be the opposite group of $G$$G$. i.e. $a{\circ }_{\mathrm{op}}b:=b\circ a$$a\circ_{\mathrm{op}} b:=b\circ a$.

Observe that $\mu :M⟼M^T$$\mu:M\longmapsto M^T$ and $\eta :M⟼M^{-1}$$\eta:M\longmapsto M^{-1}$ are group homomorphisms from $G{L}_n(R)\to G{L}_n(R{)}^{\mathrm{op}}$$GL_n(R)\to GL_n(R)^{\mathrm{op}}$,

Then the equalizer of of $\mu$$\mu$ and $\eta$$\eta$, i.e. $\mathrm{Eq}(\mu ,\eta ):=\{M\in G{L}_n(R):M^T=M^{-1}\}=O(n,R)$$\mathrm{Eq}(\mu,\eta):=\{M\in GL_n(R):M^{T}=M^{-1}\}=O(n,R)$.

Appendix The existence of equalizer in $\mathrm{Grp}$$\mathrm{Grp}$.

We know that the equalizer exists in $\mathrm{Set}$$\mathrm{Set}$. Now we need to prove that it is subgroup.

Let $f,g:G\to G^{\prime }$$f,g:G\to G'$ be two group homomorphisms, then $\mathrm{Eq}(f,g)$$\mathrm{Eq}(f,g)$ is not empty since $f(e_G)=g(e_G)=e_{G^{\prime }}$$f(e_G)=g(e_G)=e_{G'}$

Also, if $a,b\in \mathrm{Eq}(f,g)$$a,b\in\mathrm{Eq}(f,g)$, then $f(a^{-1}b)=f(a{)}^{-1}f(b)=g(a{)}^{-1}g(b)=g(a^{-1}b)$$f(a^{-1}b)=f(a)^{-1}f(b)=g(a)^{-1}g(b)=g(a^{-1}b)$. Hence $\mathrm{Eq}(f,g)$$\mathrm{Eq}(f,g)$ is a subgroup of $G$$G$.

In particular, $\ker (f)=\mathrm{Eq}(f,1)$$\mathrm{ker}(f)=\mathrm{Eq}(f,1)$, where $\mathrm{\forall }g\in G,1(g)=e$$\forall g\in G,1(g)=e$.