Answer to Question #138979 in Abstract Algebra for Sohan kumar

Question

Prove that R⁵/R² is isomorphic to R³.

Answer 1

Let $f: R^{5}\rightarrow R^{3}\\ f(x_{1},x_{2},x_{3},x_{4},x_{5})=(x_{1},x_{2},x_{3},0,0)$

One can check that f is a homomorphism .

Moreover it is onto.

Kernel(f)=$\{(x_{1},x_{2},x_{3},x_{3},x_{4},x_{5})|f(x_{1},x_{2},x_{3},x_{4},x_{5})=$(0,0,0,0,0)$\}$

=$\{(0,0,0,x_{4},x_{5})|x_{4},x_{5}\in R\}$ $\cong R^{2}$

Therefore by 1st isomorphism theorem $R^{5}/ R^{2}\cong R^{3}$