# Thread: Abelian groups

1. ## Abelian groups

Let G be an abelian group and let g,h Є G.

(i) Assuming that |g| and |h| are both finite, with hcf (|g| , |h|) = 1. Prove that

|g + h| = |g| |h|

(ii) prove that the direct sum of Zm and Zn is isomorphic to Zmn if and only if m and n are relatively prime.

any help would be appreciated. thanks

2. Originally Posted by davidmccormick
Let G be an abelian group and let g,h Є G.

(i) Assuming that |g| and |h| are both finite, with hcf (|g| , |h|) = 1. Prove that

|g + h| = |g| |h|
Let $\displaystyle |g|=n,|h|=m$. Then $\displaystyle g^n = h^m = 0$.
This means $\displaystyle (g+h)^{nm} = g^{nm}+h^{nm} = (g^n)^m + (h^m)^n = 0 + 0 = 0$
Now argue that $\displaystyle nm$ is the least such exponent for $\displaystyle g+h$.

(ii) prove that the direct sum of Zm and Zn is isomorphic to Zmn if and only if m and n are relatively prime.

any help would be appreciated. thanks
The group $\displaystyle \mathbb{Z}_n\times \mathbb{Z}_m$ has order $\displaystyle nm$. If you can show it is cyclic i.e. find a generator then by isomorphism proporties of cyclic groups it would mean that it has to be isomorphic to $\displaystyle \mathbb{Z}_{nm}$.

3. We obtain Zn x Zm by the component-wise addition of Zm and Zn. So, it is definitely cyclic with generator 1 for instance. But that only guarantees isomorphism with either Z or Zn. How can we extend that to Zmn. Also what about the reverse implication?
thanks

4. Originally Posted by davidmccormick
We obtain Zn x Zm by the component-wise addition of Zm and Zn. So, it is definitely cyclic with generator 1 for instance. But that only guarantees isomorphism with either Z or Zn. How can we extend that to Zmn. Also what about the reverse implication?
thanks
Because the order of $\displaystyle \mathbb{Z}_n\times \mathbb{Z}_m$ is $\displaystyle nm$. The group $\displaystyle \mathbb{Z}$ has infinite order therefore it cannot be the isomorphic group, while, the groups $\displaystyle \mathbb{Z}_k$ all have order $\displaystyle k$ and therefore $\displaystyle k=nm$.