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By definition:

Given an abelian group $A$, a subset $X \subseteq A$ and $x_1,\dots,x_k \in X$ with $x_i \neq x_j$ for all $1 \leq i \neq j \leq n$, the elements $x_1,\dots,x_k$ are linearly independent if they satisfy the following condition:

\begin{equation} n_1x_1 + \dots + n_kx_k = 0 \, , \, \text{with} \hspace{2mm} n_1,\dots,n_k \in \mathbb{Z} \implies n_i = 0 \hspace{2mm} \text{for all} \hspace{2mm} 1 \leq i \leq n\end{equation}

Additionally, if $X$ is a set of generators for $A$, i.e. $\langle X \rangle = A$ and if, for all $x_1,\dots,x_k \in X$, we have that $x_1,\dots,x_k$ are linearly independent, we say that $X$ is a base for $A$.

I need to prove this statement: let $A$ be an abelian group and let $X \subseteq A$ be a subset. Then $X$ is a base for $A$ iff every element $a \in A$ can be written uniquely as a finite linear combination of elements of $X$ with integers as coefficients, i.e $\exists \hspace{0.3mm} ! \, x_1,\dots,x_k \in X \, , \, n_1,\dots,n_k \in \mathbb{Z}$ so that $a = n_1x_1 + \dots + n_kx_k$.

I was able to prove the left implication $(\Longleftarrow)$ and in the right implication $(\Longrightarrow)$ proving the existence of a linear combination is trivial. However, I'm struggling to prove uniqueness. By existence, we have $x_1,\dots,x_k \in X$, $n_1,\dots,n_k \in \mathbb{Z}$ so that $a = n_1x_1 + \dots + n_kx_k$. Let $y_1,\dots,y_h \in X$, $m_1,\dots,m_h \in \mathbb{Z}$ be elements so that $a = m_1y_1 + \dots + y_hm_h$. Then we have: \begin{equation}n_1x_1 + \dots n_kx_k - m_1y_1 - \dots - m_hy_h = 0\end{equation} I've made several attempts from this step, how do we get the thesis from this? I'm pretty sure we need to use the fact that the elements in $X$ are linearly independent.

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1 Answer 1

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An arbitrary subset $X=\{x_i\}_{i\in I} \subset A$ is said to be linearly independent if $\sum_{i}n_ix_i=0\implies n_i=0$ for all $i\in I $, where $I $ is an index set.

Suppose $X $ is a base for $A $. Then any $x\in A $ can be written as $\sum_{i}n_ix_i$, where $n_i=0$ for all but finitely many $i$. Suppose there exists another representation of $x $, say $\sum_{i}m_ix_i$, defined in a similar fashion as above. Then $\sum_{i}(m_i-n_i)x_i=0$ and hence $m_i=n_i $ for all $i $.

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