# Homework Help: Abstract Algebra. Need help!

1. Sep 8, 2008

### sutupidmath

1. The problem statement, all variables and given/known data
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 $$h^{-1}$$.

3. 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.

2. Sep 8, 2008

### 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.

3. Sep 9, 2008

### 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.

4. Sep 9, 2008

### Dick

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.

5. Sep 9, 2008

### sutupidmath

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

6. Sep 9, 2008

### Dick

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

7. Sep 9, 2008

### sutupidmath

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

8. Sep 9, 2008

### 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.

9. Sep 15, 2008

### 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?

10. Sep 16, 2008

### sutupidmath

Re: Abstract Algebra. I STILL Need help on this one!

wwwww

11. Sep 16, 2008

### 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?