TITLE: Prove that $U_k(n)$ is a subgroup of $U(n)$
QUESTION [5 upvotes]: For each divisor $k\gt 1$ of $n$, let $U_k(n) = \{x \in U(n) ~|~ x\pmod {k} = 1\}$.
Prove that $U_k(n)$ is a subgroup of $U(n)$
($U(n)$ is the group formed by the positive integers less than $n$ that are coprime to $n$)

My attempt :
1) $1\equiv 1 \pmod{k}$, so $e\in U_k(n)$  
2) $a,b \in U_k(n) \implies ab\equiv 1 \cdot 1 \equiv 1 \pmod{k}$, so $ab \in U_k(n)$  
3) Since $\gcd(a,n)=1$ and $k\mid n$, we have $\gcd(a,k)=1$, for all $a\in U(n)$.
Also, since $a^{-1}\in U(n)$, we have $\gcd(a^{-1}, k) = 1$.
I'm not sure how to proceed from here. Appreciate any help. Thanks!
(I'm trying to show $a^{-1} \equiv 1 \pmod{k}$ at 3rd step )

REPLY [2 votes]: The canonical projection $U(n) \to U(k)$ is a group homomorphism with kernel $U_k(n)$.