Group generators and isomoprhisms Hello I have a rather quick question about the following;
I want to show that the groups
$\mathbb{Z_{2}} \times \mathbb{Z_{3}} \cong \mathbb{Z_{6}}$
Two questions;
I learnt the Chinese remainder theorem for rings, does it also hold for groups?
Second question
I know that any two cyclic groups with the same number of elements are isomorphic,
is the following a valid understanding of generators etc,
from $(1,1)$ since we are dealing with $\mathbb{Z}$ then our operations are addition (as 'multiplication') from addition and multiplication we can form all other 6 ordered pairs of such and therefore is that sufficient to conclude it is a cyclic group with 6 elements and clearly then isomorphic to $\mathbb{Z}$
 A: I don't know about the Chinese Remainder Theorem for groups.  But as for your second question, it is sufficient to show that $\mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic by demonstrating that $(1,1)$ is a generator.
A: About the Chinese remainder theorem, it gives a unitary ring isomorphism between (provided that $gcd(m,n)=1$) :
$$\phi:\mathbb{Z}_{nm}\rightarrow \mathbb{Z}_n\times \mathbb{Z}_m$$
On the other hand, if you have an isomorphism of rings $\psi:(A,+_A,\times_A)\rightarrow (B,+_B,\times_B)$ then this ring isomorphism gives an isomorphism of groups $\psi_+:(A,+_A)\rightarrow (B,+_B)$ defined by $\psi_+(a):=\psi(a)$ that is $\psi_+$ just "forgets" the multiplicative structure. The fact that $\psi_+$ is a group morphism is obvious from the fact that $\psi$ is an isomorphism of ring and, as a function $\psi$ and $\psi_+$ are the same functions so $\psi$ is bijective because $\psi_+$ is. Hence the Chinese remainder theorem is perfectly valid in your case.
About your second remark, it seems right but the end "isomorphic to $\mathbb{Z}$" is wrong. What you  should do is the following. 
You already highlighted a generator of $\mathbb{Z}_2\times \mathbb{Z}_3$ : $(1,1)$. In order to show the isomorphism properly define :
$$\Phi:\mathbb{Z}\rightarrow  \mathbb{Z}_2\times \mathbb{Z}_3$$
$$n\mapsto (n,n)$$
You already justified that $\Phi$ is surjective. It is clearly a group morphism. Find $Ker(\Phi)$ and then use some basic group theory to show that $\mathbb{Z}/Ker(\Phi)$ is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_3$ and conclude.
In general if $(G,.)$ is a group and you have found a generator $g_0$ of $G$ you just define :
$$\Phi:\mathbb{Z}\rightarrow G $$
$$n\mapsto g_0^n$$
The same idea allows you to justify that if $g_0$ is of infinite order then $G$ is isomorphic to $\mathbb{Z}$, otherwise $G$ is isomorphic to $\mathbb{Z}_{o(g_0)}$. 
