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In my abstract algebra study material/textbook I found the following question:

Let $S=\{a,b,c\}$. Define a suitable binary operation (if possible) $f:S \times S \to S$ such that it satisfies the following properties:

  • $f$ is associative
  • Every element in $S$ has an identity w.r.t $f$ and that identity is unique for all elements of $S$.
  • Every element in $S$ has an inverse w.r.t $f$ and that inverse is unique for that element in $S$.
  • $f$ is commutative on $S$ Also, show an example of $f$ if $S=\{0,1,2\}$.

I thought of binary operations like addition, subtraction, multiplication, division, but none of them seem to satisfy all the four conditions.

Next, I thought of defining the operation $f$ on $S$ as follows:

$$f(a,b)=f(a,b)=c;f(b,c)=f(c,b)=a;f(a,c)=f(c,a)=b;f(a,a)=a;f(b,b)=b;f(c,c)=c$$

But even this does not work as $f(a,f(b,c))=a$ while $f(f(a,b),c)=c$. Hence, $f$ is not associative in this case.

I have tried to come up with other examples but none of them satisfied all the stated properties.

So, my query is: Does such an $f$ even exist for a general $\{a,b,c\}$ or at least for $\{0,1,2\}$ ?

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  • $\begingroup$ I don't get it: the path of least resistance is just to identify $a,b,c$ as elements of, say, the abelian group $\mathbb Z_3$. That would give you an abelian group operation, which has all those properties, does it not? $\endgroup$ – rschwieb Jan 15 '18 at 15:45
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    $\begingroup$ @rschwieb What do you mean by path of least resistance? Please bear with me. I’m new to abstract algebra. Also I don’t really know what Abelian group is. Looking it up. $\endgroup$ – user400242 Jan 15 '18 at 15:47
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    $\begingroup$ When you are working with operations on small sets, write down the table of the operation, like you did in primary school for $+$ and $\times$. Since one element should be the identity you already know the result in its row and column. Since the operation is commutative, the table is symmetric with respect to the diagonal.There are no so many choices for the 4 remaining entries of the table, specially once you use that each element has an inverse. $\endgroup$ – orole Jan 15 '18 at 15:52
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    $\begingroup$ $\begin{bmatrix}f&a&b&c\\a&a&b&c\\b&b&x&a\\c&c&a&y\end{bmatrix}$ In this table I chose $a$ to be the identity. Find who should be $x$ and $y$. $\endgroup$ – orole Jan 15 '18 at 15:56
  • $\begingroup$ @Blue Oh... well you definitely should look up what an abelian group is. If you know what an abelian group is, then you'd probably have seen $\mathbb Z_3$, so that you'd know what you were aiming for. $\endgroup$ – rschwieb Jan 15 '18 at 16:12
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You treat a as 0, b as 1 and c as 2 and define the binary operation on {0,1,2} as addition module 3, under this operation it forms a cyclic group of order 3.

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  • $\begingroup$ Please define what addition module $3$ means. Also you did not talk about the general case $\{a,b,c\}$. $\endgroup$ – user400242 Jan 15 '18 at 15:42
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like if x*y = (x + y)modulo 3, and for general case, use the bijection between { 0,1,2} and { a,b,c}

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  • $\begingroup$ Thank you for the first example. What do you mean by “bijection between” $\{0,1,2\}$ and $\{a,b,c\}$ ? I don’t understand that part. $\endgroup$ – user400242 Jan 15 '18 at 15:52
  • $\begingroup$ @Blue Just google or Wiki search "bijection" $\endgroup$ – rschwieb Jan 15 '18 at 16:14

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