- #1
sutupidmath
- 1,630
- 4
Homework Statement
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].
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!