Proving Subgroups of Free Abelian Groups: A Troubleshooting Guide

In summary, the author is trying to show that if a given scalar multiple of a generator in a free abelian group is in the group, then it must be a multiple of the generator. However, he may not have been clear about what he wanted to show.
  • #1
gonzo
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I'm working on a proof for subgroups of free abelian groups and am having trouble with a step (I know other methods, but would like to try and make this one work if possible).

The basic idea is let G be a free abelian group with generators [itex](g_1...g_n)[/itex] and let H be a subgroup of G.

Assuming a suitable renumbering of the generators so that g1 does not have all zero coefficients in H, we can find a minimal coefficient of g1 in H (with respect to absolute value, and non-zero of course), and then it is easy to show that all other coefficient of g1 in H have to be multiples of this coefficient. Call it [itex]\alpha[/itex].

I realized that might not have been clear. What I mean is let:

[itex]h=a_1 g_1 +...+a_n g_n \in H[/itex]

then [itex]\alpha[/itex] is minimal of all a1 so that we know there is some element [itex]h_2[/itex] in H such that[itex]h_2=\alpha g_1 +...+a_n g_n \in H[/itex]

And all other a1's in an arbitrary h are multiples of [itex]\alpha[/itex].

What I would like to do is show that [itex]\alpha g_1 \in H[/itex]

But maybe this isn't necessarily true? Or am I missing something simple to show this?
 
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  • #2
Hrm. I still don't understand exactly what you want to say. But maybe a bad example will be useful!

Consider the free abelian group on 2 generators, G = Z².

Let H = { (m, n) | m + n is even }

H is a subgroup of Z². The smallest scalar multiple of the generator (1, 0) that lies in H is 2(1, 0). Alas, 1(1, 0) + 1(0, 1) lies in H, and |1| < |2|.
 
  • #3
I guess I wasn't clear enough, sorry. However, even though you seem to have misunderstood what I meant your example does show why I was having trouble proving my result since it is apparently not true, so thanks.

I was looking for a simplification of a proof, but I guess that was wishful thinking.
 
  • #4
what are you trying to prove?
 
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  • #5
I figured it out, but thanks anyway.
 

1. What is a free abelian group?

A free abelian group is a type of mathematical structure that consists of a set of elements and operations of addition and subtraction. It is called "free" because the elements can be combined in any way without any restrictions, and "abelian" because the order of the elements does not affect the outcome of the operations.

2. How do you prove that a group is free abelian?

To prove that a group is free abelian, you must show that it satisfies the two defining properties of a free abelian group: being free and being abelian. This can be done by demonstrating that the group has a basis, meaning that every element can be expressed as a unique combination of basis elements, and that the group is commutative, meaning that the order of operations does not matter.

3. What is the difference between a free group and a free abelian group?

A free group is a group in which the elements can be combined in any way without any restrictions, but the operations are not necessarily commutative. A free abelian group, on the other hand, has both of these properties, meaning that the elements can be combined in any way and the operations are commutative.

4. Can a non-abelian group be free abelian?

No, a non-abelian group cannot be free abelian. Since a free abelian group must have both the properties of being free and being abelian, a non-abelian group would not satisfy the second property and thus cannot be considered a free abelian group.

5. How is a free abelian group used in mathematics?

Free abelian groups are used in many areas of mathematics, including algebra, number theory, and topology. They are particularly useful for studying the properties of more complex groups and for understanding the structure of mathematical objects.

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