# Thread: homomorphism and subgroup question

1. ## homomorphism and subgroup question

Let k|n. Consider ƒ:U(n) to U(n) defined by ƒ(x)=x mod k. What is the relationship between this homomorphism and the subgroup Uk(n) of U(n).
No idea on this one. Help would be greatly appreciated.

2. Originally Posted by wutang
Let k|n. Consider ƒ:U(n) to U(n) defined by ƒ(x)=x mod k. What is the relationship between this homomorphism and the subgroup Uk(n) of U(n).
No idea on this one. Help would be greatly appreciated.
What are $U(n)$ and $Uk(n)$?

3. U(n) is defined to be the group with the numbers relative prime up to n (such that the gcd between n and an x in U(n) is 1). For example U(9)=1,2,4,5,7,8. Uk(n) is defined as the set of x's that are an element of U(n) s.t. x mod k=1. For example U7(105)={1,8,22,29,43,64,71,92}.

4. Originally Posted by wutang
U(n) is defined to be the group with the numbers relative prime up to n (such that the gcd between n and an x in U(n) is 1). For example U(9)=1,2,4,5,7,8. Uk(n) is defined as the set of x's that are an element of U(n) s.t. x mod k=1. For example U7(105)={1,8,22,29,43,64,71,92}.
Well, isn't $Uk(n)=\ker f$?

5. Let k|n. Consider ƒ:U(n) to U(k) defined by ƒ(x)=x mod k. What is the relationship between this homomorphism and the subgroup Uk(n) of U(n).
Sorry there should have been a k in the second u and not an n. So the kernel is the set of elements x in the domain that map to the identity element in the co domain. Since Uk(n) = x element of U(n) s.t. x mod k = 1, then Uk(n) would just be the kernel of f. Is this what you are saying because this seems right, but really easy for some reason.

6. Originally Posted by wutang
Let k|n. Consider ƒ:U(n) to U(k) defined by ƒ(x)=x mod k. What is the relationship between this homomorphism and the subgroup Uk(n) of U(n).
Sorry there should have been a k in the second u and not an n. So the kernel is the set of elements x in the domain that map to the identity element in the co domain. Since Uk(n) = x element of U(n) s.t. x mod k = 1, then Uk(n) would just be the kernel of f. Is this what you are saying because this seems right, but really easy for some reason.
Seems right to me, but probability that I got this wrong is not almost surely zero.