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Let $(e_1, ..., e_n)$ be the standard $\Bbb Z$-basis of $\Bbb Z^n$. Let $x_1,..., x_n$ be elements of an abelian group $G$.

I want to prove that there exists a homomorphism $f$ $:$ $\Bbb Z^n$ $\rightarrow$ $G$ such that $f$$(e_i)$ $=$ $x_i$ for all $i$ $\in$ $\Bbb N$.

Please help me..


marked as duplicate by Dietrich Burde abstract-algebra Mar 24 '17 at 19:40

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  • $\begingroup$ This question is repeated again and again, see here, and here. Please do not ask it again! $\endgroup$ – Dietrich Burde Mar 24 '17 at 19:42

Lemma: Let $\mathbb{Z}^n$ have the standard basis $\{ e_1 , ..., e_n \}$. For an abelian group $G$, and a set map $g : \{ e_1, ..., e_n \} \rightarrow G$, there is a unique homomorphism $f : \mathbb{Z}^n \rightarrow G$ such that $f(e_i) = g(e_i)$ for each $1 \leq i \leq n$.

Take a set map $g : \{ e_1, ..., e_n \} \rightarrow G$ to an abelian group $G$ and put $x_i = g(e_i)$. Define $f: \mathbb{Z}^n \rightarrow G$ by sending $(a_1, ..., a_n)$ to $\sum_{i = 1}^n a_i x_i$. In fact we must define $f$ this way since if $f : \mathbb{Z}^n \rightarrow G$ is a homomorphism such that $f(e_i) = x_i$ for $1 \leq i \leq n$ then $f \left(a_1, ..., a_n \right) = \sum_{i = 1}^n a_i f(e_i) = \sum_{i = 1}^n a_i x_i$ by the distributive property. The way we have defined $f$ indeed produces a homomorphism since \begin{align*} f(a_1 + b_1, ..., a_n + b_n) = \sum_{i =1}^n (a_i + b_i) x_i = \sum_{i = 1}^n a_i x_i + \sum_{i = 1}^n b_i x_i = f(a_1, ..., a_n) + f(b_1, ..., b_n) \end{align*} for all integers $a_1, ..., a_n, b_1, ..., b_n$.


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