Understanding Subgroups of Integers: Explained Simply

[matheinste]

Hello all.

I reluctantly ask this question because it is probably,as the text states easy, but my desire to clear this point up overides my fear of looking a fool.

I quote word for word but will use words instead of the belongs to symbol.

A subset S of the set Z of integers is a subgroup of Z if 0 is in S, -x is in S,and x+y is in S for all x and y in S. It is easy to see that a non empty subset S of Z is a subgroup of Z if and only if x-y is in S for all x and y in S.

I understand the definition and I can see that 0 is in S. I can only assume that somehow because -x ( the additive inverse of x ) is in S that this guarantees that x is in S. If S were a subgroup of Z of course -x being in the subgroup means that x was in it. But we are not assuming that S is a subgroup but testing for it.

Help!

Matheinste.

 

[radou]

Well, first of all, how "can" you see that 0 is in S? S is non empty. So a is in S. And you know that aa^-1 = e ( = 0) is in S.

 

[matheinste]

Thaks radou. I am not being deliberately awkward, I just need to completely grasp the basics which I used to find uninteresting but now regard as most interesting. As S is defined only as a subset ( non empty } and has therefore no structure defined I cannot see how we can assume anything about the set from its being non empty.

More help please.

Matheinste.

 

[radou]

As S is defined only as a subset ( non empty } and has therefore no structure defined I cannot see how we can assume anything about the set from its being non empty.

More help please.

Matheinste.

Ok, what exactly is your question? Does this bother you?

It is easy to see that a non empty subset S of Z is a subgroup of Z if and only if x-y is in S for all x and y in S.

If so, you already know that 0 is in S, from the previous post. Now, (since we're proving direction "<=", and we assumed that x-y is in S for all x and y in S), what can we "do" with 0 and a?

 

[matheinste]

Thanks radou. I think I may be a little less stupid today.

I see now that if we have at least one element in S and call it x then we have x-x=0 is in S by what we are given. Then we can have 0-x is in S and so
-x is in S and this is our inverse element and so x-(-x)=x+x is in S. This fulfils the requirements stated subgroup.

Is that OK.

Matheinste.

 

[HallsofIvy]

Thanks radou. I think I may be a little less stupid today.

I see now that if we have at least one element in S and call it x then we have x-x=0 is in S by what we are given. Then we can have 0-x is in S and so
-x is in S and this is our inverse element and so x-(-x)=x+x is in S. This fulfils the requirements stated subgroup.

Is that OK.

Matheinste.

I'm puzzled by that last statement. Why is it important that x+ x be in S?
More to the point is that if x and y are in S, then so is -y and x-(-y)= x+ y is in S.

 

[matheinste]

Thankyou HallsofIvy. I take note of what you have said. I am now happy with the answer.

Matheinste.