(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let a be in a group G, and let

[tex] H=\{ a^n: n\in Z\}[/tex]. 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 [tex] h^{-1}[/tex].

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 : [tex] h=a^{n_1} ; \ \ \ h'=a^{n_2}[/tex] where [tex]n_1,n_2\in Z[/tex] .

Now, since: [tex] a\in G=> a^{n_1},a^{n_2}\in G[/tex]

Then :

[tex] hh'=a^{n_1}a^{a_2}=a^{n_1+n_2}\in H[/tex], since [tex] n_1+n_2\in Z[/tex]. ( 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 [tex] a^{n_1} [/tex] or [tex] a^{n_2}[/tex] right? or?

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

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

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

[tex] a^{-n}=(a^n)^{-1}=h^{-1}\in H[/tex], 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' : [tex] a^{-n}=(a^n)^{-1}[/tex] 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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Abstract Algebra. Need help!

**Physics Forums | Science Articles, Homework Help, Discussion**