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**Summary::**Abstract algebra

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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- Thread starter peelgie
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I have a problem with this task. Please help.

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- #2

fresh_42

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If you need help on LaTeX, see

https://www.physicsforums.com/help/latexhelp/

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$$(Z_{31}^{*} = \{1, 2, 3, ...,30\},\cdot_{31}) $$

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fresh_42

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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

fresh_42

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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 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?

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

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What is the ##^*31## operation?

- #8

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mod 31

- #9

fresh_42

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

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