I'm stuck on the following exercise from Herstein's "Topics in Algebra":  "show that ($n$ is prime) $\Leftrightarrow ([a][b]=[0]\Rightarrow [a]=[b]=[0]) $ in $J_n$".    for the rightward implication I have:  $[a][b]=[ab]=[0]\Rightarrow n|ab\Rightarrow n|a$ or $n|b$ (or both) $\Rightarrow [a]=[0]$ or $[b]=[0]$  and I'm stuck on the leftward implication.  Why do I get $[a]=[0]$ OR $[b]=[0]$ and not $[a]=[0]$ AND $[b]=[0]$ in the leftward implication? Where is it that I go wrong?   I'd also appreciate any comment/hint about how to prove the remaining left implication.
Suppose $$ab \equiv 0 \pmod n$$ and that $a$ and $b$ are positive integers both less than $ n$. Does it follow that either $a | n$ or $b | n$?  If it does follow, give a proof.  If it doesn’t, then give an example.   For this question is this a suitable answer: No it doesn't. $3$ and $4$ are both less than $6$. $$3 \times 4 \equiv 0 \pmod {6}$$ but $4 \not \mid 6$.
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