Abstract Algebra Homework: Proving H is a Subgroup of G

[sutupidmath]

Homework Statement

Let a be in a group G, and let

H=\{ a^n: n\in Z\}. Show the following:

(i) if h and h' are in H, so is hh'.

(ii) The identity e of G is in H.

(iii) if h is in H, so is h^{-1}.

The Attempt at a Solution

Here is what i tried. First of all i am not sure that what i have done is correct, second i am not sure how a bein an element of G helps here. However i have tried to make use of such a fact in one or another way. Let's see:


(i) Let h and h' be in H, then both of them are of the form : h=a^{n_1} ; \ \ \ h'=a^{n_2} where n_1,n_2\in Z .

Now, since: a\in G=> a^{n_1},a^{n_2}\in G

Then :

hh'=a^{n_1}a^{a_2}=a^{n_1+n_2}\in H, since n_1+n_2\in Z. ( Note: This is also where i am not sure wheter i am properly making use of the fact that a is in G, because if a wouldn't be in G, then we would not be sure that we can have a^{n_1} or a^{n_2} right? or?


(ii) I am not sure on this one at all, here is what i said:

let e=a^0 be the identity element in G, so e=a^0\in H since n=0\in Z

(iii) Since h\in H then h is of the form h=a^n\in H for some n in Z. But also a^{-n}\in H since -n\in Z. So we have:

a^{-n}=(a^n)^{-1}=h^{-1}\in H, again here i think , that if a would not be an element of the group G, then we wold not be able to perform the following step, that i did on my 'proof' : a^{-n}=(a^n)^{-1} right?




I would appreciate any further and more detailed explanation from you guys. Even if i am pretty close, please try to give me further explanations.

Thanks in advance!

 

[mhazelm]

Here's what I would say:

I don't have my abstract algebra book handy, but if I remember the properties of groups well enough:

i ) looks okay to me.

ii) I would say something like: Since G is a group, if a is in G then a^{-1} is in G. Now a = a^{1} \in H and a^{-1} \in H since 1 and -1 are integers. But a*a^{-1} = a^{0}= e, thus e is in H.

iii) Similarly:

Let h be in H. Then h = a^{n} for some n in Z. Now consider h' = a^{-n}, which is in H since -n is in Z.
Now h*h' = a^{n-n} = a^{0} = e. (It's trivially true the other way, h'*h=e). Since h*h' = h'*h = e, h' is the inverse of h.

 

[sutupidmath]

Well, for part (ii) of this problem, here is what my prof said.

Since he said we have to assume that G, is finite, then it means that the elements of G will start to repeat after some time. That is

a^m=a^n for some m,n in Z. then these two el. are also in H. Also there exists (a^n)^{-1} so multiplying from the right there we will get:

a^{m-n}=e\in G and also in Z.

 

[Dick]

Well, for part (ii) of this problem, here is what my prof said.

Since he said we have to assume that G, is finite, then it means that the elements of G will start to repeat after some time. That is

a^m=a^n for some m,n in Z. then these two el. are also in H. Also there exists (a^n)^{-1} so multiplying from the right there we will get:

a^{m-n}=e\in G and also in Z.

H is a subgroup even if you don't assume G is finite. a^0 is defined as e. And a^(-n) is defined as the inverse of a^n. If you meant n is an element of N (the POSITIVE integers), then you need an additional assumption, and yes, G finite will do it.

 

[sutupidmath]

Well, i guess we needed that extra assumption about G being finite, since we haven't learned subgroups yet, so i guess we couldn't make use of that fact, that H is a subgroup of G.

 

[Dick]

Well, i guess we needed that extra assumption about G being finite, since we haven't learned subgroups yet, so i guess we couldn't make use of that fact, that H is a subgroup of G.

I guess the real question is what did you mean by Z? Is it all integers, or only the positive ones?

 

[sutupidmath]

I guess the real question is what did you mean by Z? Is it all integers, or only the positive ones?

Well, in the book they did not specify it, but i guess they meant all integers.

 

[Dick]

Then you don't need that G is finite. I probably shouldn't have said "subgroup". But the properties you are trying to show imply H is a subgroup of G.

 

[sutupidmath]

How would one prove that if h is in H, then also h^-1, that is its inverse of it is also in H, if Z is only positive integers.

I managed to show, as stated that e=a^{m-n}\in H since i supposed that the group G is finite, and i also know that now i have to take two elements in H, say h and h' and then use the property that H is closed, and

hh'=...=e\in H, but i cannot figure out how to pick up h and h'?

Can u help me on this?

 

[sutupidmath]

wwwww

 

[sutupidmath]

I think i got it, let's see:

SInce g^{m-n}=e\in H it means that H consists of m-n elements, that is all powers of g up to m-n. Now this means that there defenitely are:

s_1, s_2 \in Z^{+} such that s_1+s_2 =m-n while [/tex] g^{s_1},g^{s_2} \in H[/tex] this means that:

For every element in H, say h=g^{s_2}, s_2 \in Z^+, \exists h'=g^{s_1}, s_1 \in Z^+ such that

using closure property of H, we get:


hh'=g^{s_1}g^{s_2}=g^{s_1+s_2}=g^{m-n}=e \in H which means that h and h' are inverses of each other, and still both h and h' are in H. Proof done. RIght?