Does the multiplicative Inverse have to be co prime with the module

Question

So I am working on a proof right now, that one of my friends gave me and it says, that, when I have a multiplicative inverse of a that that mutliplicative inverse is automatically co prime to the module if gcd(a,m) = 1.

So that the multiplicative Inverse of a is automatically co prime to m

I just do not get the math behind that.

Thanks

Answer 1

Let $ gcd(a,m) = 1$ and $ ab \equiv 1 ~mod ~m$. Then $ ab = 1 + rm, \ r\in \mathbb{Z}$.

If $ x = gcd(b,m) $, it follows that $ x | ab $ and $ x|rm $. Consequently, $ x| ab - rm \implies x |1 \implies x=1 $.