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 as follows: $$\begin{matrix}{*}&{0}&{1}&2&3\\{0}&{0}&{1}&2&3\\{1}&{1}&{2}&3&0 \\ {2}&{2}&{3}&0&1\\ {3}&{3}&{0}&1&2 \end{matrix}$$. The identity element of the group S is determined to be 0, as it satisfies the condition x*0=0*x for all x in S. The inverses for each element in S are also provided, showing that 0 is its own inverse, while 1, 2, and 3 have inverses of 3, 2, and 1 respectively.

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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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