Linearly Independent Sets in Abelian Groups

[Bashyboy]

Homework Statement

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$X$ is linearly independent if and only if every nonzero element of the subgroup $\langle X \rangle$ may be written uniquely in the form $n_1 x_1 + ... n_k x_k$ ($n_i \in \Bbb{Z} \setminus \{0\}$, and $x_1,...,x_k \in X$ are distinct).

Homework Equations

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A subset $X$ of an abelian group $F$ is said to be linearly independent if $n_1 x_1 + ... + n_k x_k$ always implies $n_i = 0$ for every $i$ (where $n_i \in \Bbb{Z}$ and $x_1,...,x_k$ are distinct elements in $X$.

The Attempt at a Solution

I will focus on the $\impliedby$ direction. Here is a solution I found online which I feel is erroneous:

"Given $\sum n_i x_i = \sum m_i x_i$, we get $\sum (n_i - m_i)x_i = 0$ If every element of $\langle X \rangle$ is the unique linearly combination of elements from $X$, then to begin with we know $n_i = m_i$ for each $i \in I$ Since $0 = \sum 0x_i$, it follows any linear combination $\sum n_i x_i = 0$ forces $n_i = 0$ for all $i \in I$; therefore, $X$ is linearly independent. "

This seems wrong. First, they are assuming that *every* element in $\langle X \rangle$, including the identity $0$, can be uniquely written as certain combination of elements in $X$; but this is not what the problem statement says. Second, it doesn't seem that we can infer that $n_i = 0$ for every $i$ from $\sum n_i x_i = \sum 0 x_i$, since $0$ can (possibly) be written in a variety of ways and we are *assuming* that $\sum n_i x_i$ is zero.

My thought was to do something like the following. Suppose every $x \in \langle X \rangle \setminus \{0\}$ can be uniquely represented in the manner already mentioned. Let $x_1,...,x_n$ be distinct, nonzero elements in $X$, and suppose $k_1 x_1 + ... + k_n x_n = 0$. If $k_n x_n \neq 0$, then $k_1 x_1 + ... k_{n-1} x_{n-1} = 0x_1 + ... 0x_{n-1} -k_n x_n$. If the RHS is zero, then we are done. If not, then we have two different representations of the same element, unless $n_i = 0$ for every $i$. Now, if $k_n x_n = 0$, then choose the next nonzero element.

This seems incorrect, too, and, as you may have noticed, I am implicitly assuming that every element has infinite order. What's going on? How do we prove the statement?

 

[andrewkirk]

Regarding an implicit assumption that all elements have infinite order: I don't know exactly how the vector space concept of 'linearly independent' is applied to abelian groups, but if it is applied with exactly the same definition as for vector spaces then a linearly independent set cannot contain any cyclic elements. So the implicit assumption is fine. I am not sure it's needed though.

Let's state, for definiteness, the presumed definition of LI for abelian groups. A set X of elements of an abelian group G is LI iff for any finite subset $\{x_1,...,x_n\}$ of X and set of integers $\{k_1,...,k_n\}$ such that $\sum_{j=1}^n k_jx_j=0$ it must be the case that all $k_i$ are zero. Note that that immediately implies that X does not contain zero or any cyclic elements.

Now assume that there exist $\{x_1,...,x_n\}$ of X and integers $\{k_1,...,k_n\}$ such that $\sum_{j=1}^n k_ix_i=0$. Arbitrarily choose $j\in\{1,...,n\}$. Then we can write:
$$-k_jx_j = \sum_{\substack{l=1\\l\neq j}}^n k_lx_l$$

Do you think you can take it from there?