Prove that this Function is a Homomorphism

  • Thread starter peelgie
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  • #1
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Summary:: Abstract algebra

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I have a problem with this task. Please help.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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  • #3
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$$(Z_{31}^{*} = \{1, 2, 3, ...,30\},\cdot_{31}) $$
 
  • #4
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You have to show some efforts so that we can see where your problems are. We will not do the homework for you.
 
  • #5
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(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?
 
  • #6
13,457
10,517
(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.
 
  • #7
LCKurtz
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What is the ##^*31## operation?
 
  • #8
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mod 31
 
  • #9
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mod 31
Yes, sure. But why is it a ring homomorphism?
 

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