Question

Prove that R ^n/R^m~=R^n-m as groups, where n, m belongs to N such that n>=m.

Answer 1

Answer to Question #82609, Math / Abstract Algebra

Question. Prove that $\mathbb{R}^n / \mathbb{R}^m \cong \mathbb{R}^{n - m}$ as groups, where $n, m$ belongs to $\mathbb{N}$ such that $n \geq m$.

Answer. Define a projection group homomorphism $f: \mathbb{R}^n \to \mathbb{R}^{n - m}$ by

$$f \left(\left(a _ {1}, \dots , a _ {m}, a _ {m + 1}, \dots , a _ {n}\right)\right) = \left(a _ {m + 1}, \dots , a _ {n}\right).$$

We will use the First Group Isomorphism Theorem.

- The kernel of $f$ is the set of all elements $a$ of $\mathbb{R}^n$ having the form $(a_1, \ldots, a_n)$ such that all of the equivalent conditions below hold:

- $f((a_1, \ldots, a_m, a_{m+1}, \ldots, a_n)) = (0, \ldots, 0)$;

- $(a_{m + 1},\ldots ,a_n) = (0,\ldots ,0)$;

- for all $i \in \{m + 1, \dots, n\}$, $a_i = 0$;

- $a$ has the form $(a_1, \ldots, a_m, 0, \ldots, 0)$.

Clearly, $\ker f$ has dimension $m$ and is $\mathbb{R}^m$ viewed as a subgroup of $\mathbb{R}^n$.

- For every $(a_{m + 1},\ldots ,a_n)\in \mathbb{R}^{n - m}$,

$$f \left(\left(0, \dots , 0, a _ {m + 1}, \dots , a _ {n}\right)\right) = \left(a _ {m + 1}, \dots , a _ {n}\right).$$

Hence $f$ is surjective, and the range (image) of $f$ is $\mathbb{R}^{n - m}$.

By the First Group Isomorphism Theorem, $\mathbb{R}^n / \mathbb{R}^m \cong \mathbb{R}^{n - m}$.

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