# Let $G = {5, 15, 25, 35}$. Prove that $G$ is a group under multiplication modulo $40$. (Without using Cayley Table) [duplicate]

Let $G = {5, 15, 25, 35}$. Prove that G is a group under multiplication modulo $40$. I know how to solve for associativity and closure but I don't know how to find the identity for the set. All the answers that I've seen are using the Cayley Table but I haven't learned that in school yet. Can someone please help me with another way to solve this? Any help will be appreciated.

## marked as duplicate by Chris Culter, Lord Shark the Unknown, Krish, Claude Leibovici, Patrick StevensSep 14 '17 at 7:16

• An identity solves $e^2=e$. Which of your elements solves that? – Lord Shark the Unknown Sep 14 '17 at 4:24
Consider the group $U(n)$ of units modulo $n$ under multiplication. For $m\in U(n)$ , $mU(n)$ is a group under multiplication mudulo $mn$. The identity elelment of $mU(n)$ is $m\beta$ where $\beta =m^{-1}$ in $U(n)$.