$ a\in \Bbb Z_n$ is invertible $\,\Rightarrow\gcd (a,n) =1$ I need help with this:
If a is an element of $\mathbb Z_n$ and $\gcd(a,n) > 1$ , then a is not invertible. 
First you show that if $a$ is an element of $\mathbb Z_n$ and $\gcd(a,n)>1$, then there 
is an element $b$ of $\mathbb Z_n$ and ($b$ is not equal to the zero element) from
which $ab = 0$. 
The second part I cant get. i.e. show if $b$ is not the zero element 
and $ab=0$ then $a$ is not invertible.
 A: It is the special case $\,\color{0a0}{b = 1}\,$ of the following solvability criterion.
$\qquad \exists\, x\!:\ ax\equiv b\pmod{\! n}\iff \exists\, x,y\!:\ ax+ny = b\color{#c00}{\overset{\rm Bezout\!\!\!\!}\Longleftarrow}\!\color{}{\Longrightarrow}\, \gcd(a,n)\mid b $
Remark $\ $ The OP only requires the simpler black direction $\,(\Longrightarrow),\,$ not the deeper reverse arrow $(\color{#c00}\Longleftarrow)$ using $\rm\color{#c00}{Bezout's}\,$ identity for the gcd.
A more general ring-theoretic understanding comes from viewing it as a special case of the fact that zero-divisors are never invertible (in nonzero-rings).
A: Assume $ab = 0$ and at the same time that $a$ is invertible. That means that there exists an $a^{-1}$ such that $a^{-1}a = 1$ But then we have
$$
b = 1\cdot b = a^{-1}ab = a^{-1}\cdot 0 = 0
$$
so there cannot be a $b \neq 0$ such that $ab = 0$. That means that if there is such a $b$, then $a$ cannot be invertible.
A: Suppose gcd(a,n)=1
We can write this as: ax+ny=1
And class of ax+ny is equal to class of 1
cl(ax+ny)=cl(1)
cl(ax)+cl(ny)=cl(1)
We know about class of zero i.e (cl(0)={0+ny\y belongs to Z})
From this we can write:
cl(ax)+cl(0)=cl(1)
cl(ax)=cl(1)
cl(a).cl(x)=cl(1)
cl(a)^-1=cl(x)
So cl(a) is Invertible
Conversely,
Suppose that cl(a) is Invertible
Then there exist an element cl(x) of Zn other than cl(0) such that
cl(a).cl(x)=cl(1)
cl(ax)=cl(1)
ax=1 mod(n)
n|(ax-1)
ax-1=ny (y belongs to Integers)
ax-ny=1
ax+n(-y)=1
gcd(a,n)=1
