So, in our notes, we had an example where we defined f: Z / 6 Z by g([a]) = [5a] (where z is set of all integers)
Already, I don't follow what the g([a]) = [5a] means, I'm assuming they are equivalence classes?
then he wrote :
[0]-> [0] [1] -> [5] [2] -> [4] [3] -> [3] [4] -> [2] [5] -> [1]
NOT SURE what this means either.
Finally he wrote gcd(5,6) = 1 5s + 6t= 1
not sure why this is relevant or how he used it?
Please help. He went on to prove this, but I just want to understand what is going on, or what I'm supposed to try and prove.
Yes, equivalence classes, of the equivalence relation $a\equiv b \pmod6$ which is defined as $$ 6\ \ |\ \ a-b\,.$$ In other words, in the quotient $\Bbb Z/6\Bbb Z$ we regard two numbers equal if they differ by $6k$ for some $k\in\Bbb Z$. In symbols it is just $$[a]=[b] \ \iff\ a\equiv b\pmod6\,.$$ Now observe that $$g([2])=[5\cdot 2]=[10]=[4]$$ because $10-4$ is divisible by $6$.
You can also observe that $[5]=[-1]$ which makes the calculations even easier..
The fact that $5$ is coprime to $6$, that is, $\gcd(6,5)=1$, can be just a side note at this stage. Its importance in general is that the function $x\mapsto a\cdot x$ in $\Bbb Z/m\Bbb Z$ is bijective (one-to-one correspondence) iff $a$ is coprime to $m$.
In other words, one can thus divide by $a$ modulo $m$.