Answer to Question #166667 in Linear Algebra for Sarita bartwal

Question

Apply the fundamental theorem of homomorphism to prove that

R^4/R^2 is isomorphic to R^2

Answer 1

Consider the groups $(\mathbb R^4, +)$ and $(\mathbb R^2, +)$, and define the map $\psi: \mathbb R^4 \to\mathbb R^2, \ \psi(x_1,x_2,x_3,x_4)=(x_1,x_2).$

Since $\psi((x_1,x_2,x_3,x_4)+(y_1,y_2,y_3,y_4))=\psi(x_1+y_1,x_2+y_2,x_3+y_3,x_4+y_4)= (x_1+y_1, x_2+y_2)=(x_1,y_1)+(x_2,y_2)=\psi(x_1,x_2,x_3,x_4)+\psi(y_1,y_2,y_3,y_4),$

we conclude that $\psi$ is a homomorphism.

Let us find kernel:

$\ker(\psi)=\{(x_1,x_2,x_3,x_4)\in\mathbb R^4\ |\ \psi(x_1,x_2,x_3,x_4)=(0,0)\} = \{(x_1,x_2,x_3,x_4)\in\mathbb R^4\ |\ (x_1,x_2)=(0,0)\} =\{(0,0,x_3,x_4)\ |\ x_3,x_4\in\mathbb R\}.$

The bijective map $f: \mathbb R^2 \to \ker(\psi),\ f(a_1,a_2)=(0,0,a_1,a_2),$ is an isomomorphism. Indeed, $f((a_1,a_2)+(b_1,b_2))=f(a_1+b_1,a_2+b_2)= (0,0,a_1+b_1, a_2+b_2)=(0,0,a_1,a_2)+(0,0,b_1,b_2)=f(a_1,a_2)+f(b_1,b_2).$

Taking into account that $Im(\psi)=\mathbb R^2,\ \ker(\psi)\cong\mathbb R^2,$ and according to  the fundamental theorem of homomorphism $\mathbb R^4/\ker(\psi)\cong Im(\psi)$, we conclude that $\mathbb R^4/\mathbb R^2\cong \mathbb R^2.$