# Need help! Inverse!

1. 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.
Where $$H=\{a^n: n\in Z^+\}$$ a in G
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?

Last edited: Sep 15, 2008
2. Sep 16, 2008

### tiny-tim

Hi sutupidmath!

(I'm not sure I understand what G and H are, but in case I've got it right:)

Hint: if G is finite, and Hh = {hn : n ε Z+}, then how large is Hh?

3. 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?

4. Sep 17, 2008

### tiny-tim

Hi sutupidmath!

This is very confusing …

I take it you're using g instead of a? …

and I don't understand what m is, or why you're using m - n when n is just an index.

Can't you shorten it, by starting with something like "for any s2 ε Z+, define s1 = -s2 mod m - n …" ?