Proving Torsion Subgroup Normality and Torsion-Free Property in Abelian Group G

[ehrenfest]

Homework Statement

Prove that the torsion subgroup T of an abelian group G is a normal subgroup of G, and that G/T is torsion free.

Homework Equations

The Attempt at a Solution

The second part of this exercise makes absolutely no sense to me. We know nothing about G, so why is there any reason that the nonidentity elements of G/T would all have infinite order. G could even be finite. Is the statement of the question correct? Should G be a torsion free group?

 

[Mystic998]

I think G has to be finitely generated. But I could be mistaken.

Edit: Never mind. I'm wrong.

 

[Mathdope]

If G is abelian every subgroup is normal. I assume that you mainly are trying to show that it is a subgroup.

So, if you have two elements of finite order is their product of finite order, and is the inverse of such an element of finite order? If so they form a subgroup of G.

There's nothing wrong with the statement. As Mystic998 points out, if G is finite then T = G, so that G/T = (e). It doesn't violate the definition of torsion-free.

 

[Mystic998]

Oh, if G is finite, I think G = T and G/T is torsion free trivially.

Sorry about all the replies. I'm thinking about a lot of different problems at once, and my brain is kind of jumping from topic to topic.

 

[ehrenfest]

If G is abelian every subgroup is normal. I assume that you mainly are trying to show that it is a subgroup.

So, if you have two elements of finite order is their product of finite order, and is the inverse of such an element of finite order? If so they form a subgroup of G.

There's nothing wrong with the statement. As Mystic998 points out, if G is finite then T = G, so that G/T = (e). It doesn't violate the definition of torsion-free.

Umm...how do you know that T=G if G is finite? 2 Z_10 is a torsion subgroup of the abelian group Z_10, but 2 Z_10 is not equal to Z_10.

 

[Mathdope]

As I recall the definition of the torsion subgroup is the set of all elements that have finite order. If G is finite, all elements have finite order so G = T. What's your definition?

Also, I don't know what you mean by "a" torsion subgroup. According to the defintion above (if it's what you're using) the torsion subgroup is unique.

 

[Mystic998]

Well, I could see it being a torsion Z-submodule or something. But I'm pretty sure the definition of the torsion subgroup of an abelian group G is the torsion submodule of G considered as a Z-module, which means you take all the elements that can be multiplied by a nonzero element of the integers to get zero.

 

[ehrenfest]

My book contains the following sentence:

"A torsion group is a group all of whose elements have finite order."

http://en.wikipedia.org/wiki/Torsion_group

By that definition 2 Z_10 is a torsion subgroup of Z_10.

EDIT: I see now. The term "torsion subgroup" is not merely a subgroup that is torsion, but is defined as the set of all elements of G that have finite order.

 

[Mystic998]

It's a torsion group that's a subgroup of Z_10. But it's not the torsion subgroup of Z_10.

 

[Mathdope]

EDIT: I see now. The term "torsion subgroup" is not merely a subgroup that is torsion, but is defined as the set of all elements of G that have finite order.

Right. So any abelian group has a unique torsion subgroup (assuming it's a subgroup, which they are asking you to show).