Abstract algebra question (1) order problem (2) ring problem Let G be an abelian group and let a, b ∈ G have orders m and n respectively.
Suppose furthermore that gcd(m, n) = 1. Show that ab has order mn.
I noticed that if gcd(m,n) is not 1, then order of ab is smaller than mn. 
Let R be any finite ring with unity. Show that every nonzero element of R is either a unit or a zero divisor
For this question, I am not sure how to start.
I guess for any a in R, we want to show either ab=1 or ab=0.
 A: For problem (1):
Let $|ab|=t$, then $(ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e.$ Thus $t$ divides $mn$. So $t \leq mn$. Furthermore $(ab)^t=e$ implies $a^t=b^{-t}$. Thus $|a^t|=|b^{-t}|$. But
$$|a^t|=\frac{m}{\gcd(m,t)}=\frac{n}{\gcd(n,t)}=|b^{-t}|.$$
This last expression says:
$$m | (n\gcd(m,t))$$ 
but $\gcd(m,n)=1$, so $m | \gcd(m,t)$. This implies $t=mk$ for some $k \in \mathbb{Z}$. Likewise $t=np$ for some $p \in \mathbb{Z}$. But $\gcd(m,n)=1$ implies $mn | t$. Thus $mn \leq t$. Hence $t=mn$.
For problem (2):
Let $R=\{a_1,a_2, \ldots ,a_n\}$. Let $a_1 \in R$ such that $a_1 \not \in \{ 1_R,0_R\}$. Now consider the map $x \to a_1x$. 
If this is not injective then $\exists x,y \in R$ such that $x \neq y$ and $a_1x=a_1y$. But this gives $a_1(x-y) = 0$. Hence $a_1$ is a zero-divisor.
If this is injective (hence surjective as well due to finiteness) then $\exists x \in R$ such that $a_1x=1$. Then $a_1$ is a unit, 
A: For second question,
Let $a\in R$ think $a,a^2,a^3...$. As $R$ is finite.
Edit (As AnuragA suggest)
We must have $a^i=a^j\implies a^j(a^{i-j}-1)=0$, if $a^{i-j}\neq 1$ then $a$ is zero divisior.
