# Prove that this Function is a Homomorphism

Summary:: Abstract algebra

[Moderator's note: Moved from a technical forum and thus no template.]

Last edited by a moderator:
Delta2

Related Calculus and Beyond Homework Help News on Phys.org
$$(Z_{31}^{*} = \{1, 2, 3, ...,30\},\cdot_{31})$$

fresh_42
Mentor
You have to show some efforts so that we can see where your problems are. We will not do the homework for you.

(1)
I need to prove this equation:
$$\varphi(x\cdot_{31}y) = \varphi(x)\cdot_{31}\varphi(y)$$
So:
$$\varphi(x\cdot_{31}y) = (x\cdot_{31}y)^{18} = x^{18}\cdot_{31}y^{18} = \varphi(x)\cdot_{31}\varphi(y)$$ That is Correct? Function is homomorphism?

fresh_42
Mentor
(1)
I need to prove this equation:
$$\varphi(x\cdot_{31}y) = \varphi(x)\cdot_{31}\varphi(y)$$
So:
$$\varphi(x\cdot_{31}y) = (x\cdot_{31}y)^{18} = x^{18}\cdot_{31}y^{18} = \varphi(x)\cdot_{31}\varphi(y)$$ That is Correct? Function is homomorphism?
This is correct, but it could be that you have to justify the equation in the middle: ##(x\cdot_{31}y)^{18} = x^{18}\cdot_{31}y^{18}##. It depends on what you may use and what not. Since you haven't told us this information, we cannot know.

I mean "trivial" is also a valid answer. It all depends on what can be assumed as given and what cannot.

LCKurtz
Homework Helper
Gold Member
What is the ##^*31## operation?

mod 31

fresh_42
Mentor
mod 31
Yes, sure. But why is it a ring homomorphism?