$k\in \mathbb Z,\ (k,p)=1$ the element $k\cdot1_A$ is invertible

Question

Let $A$ be a ring. The order of $1_A$ in (A,+) is p (prime). For $k\in \mathbb Z,\ (k,p)=1$ the element $k\cdot1_A$ is invertible. I tried to prove this.

$(k,p)=1 \to \exists m,n\in \mathbb Z\ s.t \ \ mk+np=1$. I don't know what can I do now.

Answer 1

You're on the right track.

Since $(k\cdot1_A)(m\cdot1_A)=km\cdot1_A=(1-np)\cdot1_A=1\cdot1_A-np\cdot1_A=1_A-n0_A=1_A,$

$(k\cdot1_A)$ has an inverse in $A$, namely, $m\cdot1_A.$