MHB Cayley table (Asnate's 's question at Yahoo Answers)

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The discussion focuses on constructing the Cayley table for the group S={0,1,2,3} under the operation defined as a*b=a+b(mod4). The Cayley table is presented, showing the results of the operation for all pairs of elements in S. The identity element of the group is identified as 0, since it satisfies the condition x*0=0*x for all x in S. Additionally, the inverses for each element are provided, with each element's inverse ensuring that their product yields the identity element. This explanation clarifies the steps involved in constructing the Cayley table, identifying the identity, and finding inverses.
Fernando Revilla
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Here is the question:

Let group S={0,1,2,3} and define a*b=a+b(mod4)
Write down the Cayley Table for (S,*)
Find the identity element of S.
Find the inverse of each element of S.

Please explain how and why you do each step. Many Thanks

Here is a link to the question:

Cayley Table construction? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Asnate,

Previous examples: $(a)\;1+2=3$ and the remainder of the euclidean division $3:4$ is $3$ so, $1*2=3$. $(b)\;2+2=4$ and the remainder of the euclidean division $4:4$ is $0$ so, $2*2=0$. $(c)\;2+3=5$ and the remainder of the euclidean division $5:4$ is $1$ so, $2*3=1$ etc. Then, you'll easily verify that the Cayley table is: $$\begin{matrix}{*}&{0}&{1}&2&3\\{0}&{0}&{1}&2&3\\{1}&{1}&{2}&3&0 \\ {2}&{2}&{3}&0&1\\ {3}&{3}&{0}&1&2 \end{matrix}$$ The element $0$ satisfies $x*0=0*x$ for all $x\in S$, hence $0$ is the identity element of $S$.

If $x\in S$, its inverse is the element $x'\in S$ satisfying $x'*x=x*x'=0$. Easily you'll find the inverse of any element of $S$: $$\begin{matrix}{x}&{0}&{1}&2&3\\{x'}&{0}&{3}&2&1 \end{matrix}$$

P.S. I've tried the $\LaTeX$ tabular environment, but it doesn't work.
 
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