# Thread: Subgroup of abelian group

1. ## Subgroup of abelian group

There seems to be little literature on the subject of abstract algebra, to help confused persons like myself. My lecturer is also the worst person to teach it, cause for the first time in my experiences of life i am entirely clueless about a subject despite my efforts.. I usually figure things out after awhile if i don't get it one time, but this subject is really trying. I got Linear Algebra kinda ok and discrete maths as well, but idk about abstract. any light shed onto this will help me a whole lot:

Let G be an abelian group, and let n be a fixed positive integer. Show that
N = {g in $\displaystyle G|g=a^{n}$ for some a in G}

2. ## Re: Subgroup of abelian group

Originally Posted by Ife
Let G be an abelian group, and let n be a fixed positive integer. Show that
N = {g in $\displaystyle G|g=a^{n}$ for some a in G}
I think the question is
let G be abelian group with finite order prove that
$\displaystyle N = \{g \in G \mid g = a^n\;\; \text{for some a in G}\}$

is a subgroup of G

first the identity is in N since $\displaystyle e^n = e$
now let g in N so $\displaystyle g = a^n \;\; \text{for some a in G}$
but a in G and G is group so $\displaystyle a^{-1} \in G$
you can see that $\displaystyle (a^{-1})^na^n = e$
so $\displaystyle (a^{-1})^n = g^{-1}$ and $\displaystyle g^{-1}$ in N since $\displaystyle g^{-1} = (a^{-1})^n$

let g,h in N so $\displaystyle g = a^n \;\;\text{for some a in G}$ and $\displaystyle h = b^n \;\;\text{for some b in g }$

$\displaystyle gh = a^n b^n = (ab)^n \in N$
so N is a subgroup