- #1
Bashyboy
- 1,421
- 5
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?