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

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.

• What is $(k\cdot 1_A)(m\cdot 1_A)?$ – J. W. Tanner Mar 10 at 18:01
• It is $km$ .... – Gaboru Mar 10 at 18:06
• It's $km\cdot 1_A$, and use $mk=1-np$ – J. W. Tanner Mar 10 at 18:07
• But the + and . are not the same from "$mk+np=1$"(these are the standard + and . from Z) – Gaboru Mar 10 at 18:10
• That's why I'm confused – Gaboru Mar 10 at 18:11

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, $$m\cdot1_A.$$