# Is a single element linearly independent in an abelian group?

This is my definition of linearly independent elements in an abelian group:

Let $$A$$ be an abelian group, let $$X \subseteq A$$ be a subset 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 defined to be linearly independent if they satisfy the following condition:

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

I'm proving theorem 1.6 of Hungerford's Algebra and got stuck at "Either $$G \cap H = \{0\}$$, in which case $$G = \langle d_1x_1 \rangle$$ and the theorem is true [...]". Why is $$d_1x_1$$ linearly independent? I need this in order to say that $$\{d_1x_1\}$$ is a basis of $$G$$. I'm pretty sure this comes from the fact that $$d_1x_1 \neq 0$$, but linearly independence does not seem to be true because $$nd_1x_1 = 0$$ might be true even though $$n \neq 0$$. For example, if $$d_1x_1$$ is an element of order $$n$$. So, is the implication $$$$n(d_1x_1) = 0 \quad \wedge \quad d_1x_1 \neq 0 \quad \implies \quad n = 0$$$$ true or false in the context of abelian groups? What if we consider free abelian groups instead? If it's always false, then how is the linearly independence proven in Hungerford's proof?

• What is $G$ and $H$? – Wuestenfux Sep 17 '19 at 15:21
• $G < F$ is a subgroup of a free abelian group, $H$ is defined as a particular subgroup of $G$. But that's not the point, I just don't understand why a non-zero element is automatically linearly independent. Is this even true? – Lele99_DD Sep 17 '19 at 15:26
• Is $A$ an arbitrary abelian group, or a free abelian group? (More specifically, does this definition apply to general abelian groups, or only to free abelian groups?) – Morgan Rodgers Sep 17 '19 at 16:35
• @MorganRodgers The definition applies to general abelian groups. – Lele99_DD Sep 17 '19 at 17:26
• But Theorem 1.6 (now that I see it posted in an answer below) is specifically pertaining to free abelian groups. In this case no element has finite order. You are correct that a single element that has finite order is not linearly independent. – Morgan Rodgers Sep 17 '19 at 18:47

You’re correct that, for general abelian groups, any set containing finite order elements is not linearly independent.

For example, in $$\mathbb{Z}/4\mathbb{Z}$$ I can take $$n = 2$$ and $$x_1 = 2$$. Then neither element is zero, but $$2\cdot2 = 0$$. Thus $$\{2\}$$ is not linearly independent.

However, in the free abelian case, this fails. If $$nx_1 = 0$$ for $$n\ne 0$$ then this implies $$x_1$$ has finite order. No nontrivial element of any free abelian group has finite order, so this won’t occur in your case.

Let me address your question as formulated in your comment, namely why a single nonzero element of a free abelian group automatically forms a linearly independent subset.

What I think you might be missing is that the summation operator $$\sum_{i=1}^k s_i$$ is defined just as well for $$k=1$$ as it is for $$k \ge 2$$. In fact, in a rigorous treatment, one would define the $$k$$-ary summation operator of any associative binary operation by induction on $$k \ge 1$$. The basis step of the induction is to define $$\sum_{i=1}^1 s_i = s_1$$. In the inductive step, assuming that $$k \ge 2$$ and that $$\sum_{i=1}^{k-1} s_i$$ is already defined, one defines $$\sum_{i=1}^{k} s_i = \left(\sum_{i=1}^{k-1} s_i\right) + s_{k}$$.

All that goes to say that the definition of linearly independence is about linear combinations, which are just special kinds of finite sums, and a sum of one object is allowed as part of this definition:

A subset of $$k$$ distinct elements $$\{x_1,...,x_k\} \subseteq A$$ (where $$k \ge 1$$) is said to be linearly independent if for any sequence of integers $$(a_1,...,a_k)$$, if $$\sum_{i=1}^k a_i x_i=0$$ then $$a_i=0$$ for each $$1 \le i \le k$$.

• In the case of finite order elements, shouldn't the correct definition (as OP discusses) be that $\sum_{i=1}^k a_ix_i=0$ implies that each $a_ix_i=0$? – WoolierThanThou Sep 17 '19 at 15:56
• While I agree that the original question seems not to have the "free abelian" requirement, I will restrict my answer to the free abelian situation as the question was reformulated in the comment. It's not clear to me that the concept of "linear independence" makes much sense in a non-free module, although I'm happy to be corrected. – Lee Mosher Sep 17 '19 at 15:59
• Sorry, maybe I was unclear, but this is not what I was looking for. I'm fine with $\sum_{i=1}^1 s_i$ being equal to $s_1$ and I understand that the definition of linearly independence works in the case $k=1$, too. My problem is just the implication $a_1x_1 = 0 \wedge x_1 \neq 0 \implies a_1 = 0$. – Lele99_DD Sep 17 '19 at 16:06
• I have a feeling that the implication $a_1x_1 = 0 \wedge x_1 \neq 0 \implies a_1 = 0$ is true in free abelian groups, but not in arbitrary abelian groups because of the fact that every element in a free abelian group has infinite order, whereas abelian groups might have finite-order elements. But this is just a conjecture to me, I'm not sure whether this is true or not. – Lele99_DD Sep 17 '19 at 16:12
• Alright. It would help, then, if you edited your question to make clear what you are and are not asking, particularly given the direction of your comment. – Lee Mosher Sep 17 '19 at 16:17

According to the Hungerford which you seems to try reading, $$x_1$$ is choosen so that $$W=\{x_1, y_2, \cdots, y_n \}$$ is a basis of $$F$$.

(Here $$x_1-y_1$$ is a linear combination of $$y_2, \cdots, y_n$$, where $$\{y_1, y_2, \cdots, y_n \}$$ is a basis of $$F$$ selected tricky. For details, see the proof above.)

Observe that $$nd_1 x_1= nd_1x_1 + 0y_2 + \cdots + 0y_n$$. As a result, $$nd_1x_1=0$$ implies $$nd_1=0$$.