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In the problems involving two algebraic systems, for eg.,$\langle S,*\rangle$ and $\langle P,\bigoplus\rangle$ where the sets $S=\{a,b,c\}$ and $P=\{1,2,3\}$. Here we have to check whether they both are isomorphic or not. While solving, they take values as $g(a)=3$, $g(b)=1$ and $g(c)=2$ and prove the systems as isomorphic. If I try other combination of values, it doesn't satisfy isomorphism. Then, on what basis these values are chosen(a=3,b=1,c=2)?

Kindly check out this Pg. 234 for the definitions of the operations.

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    @MartinSleziak Ok sir. Thank you.2012-08-23

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Look at the multiplication tables at the bottom of the page in your link, and try rewriting the second one with the columns and rows in the order $3,1,2$ instead of $1,2,3$. You should see something that looks almost identical to the table on the left, but with different symbols. Specifically, $a$ is replaced by $3$, $b$ by $1$ and $c$ by $2$. This is why the two are isomorphic - the two algebraic structures are the same, just the symbols for the elements are different. If you reshuffle the columns on the right-hand side in any other way, the tables won't match up properly, which is why other definitions of $g$ won't work.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/46$0$3/discussion-between-gomathi-and-matt-pressland)2012-08-23
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You will have to check to see if $g(a\ast b)=3\oplus 1 \\ g(b\ast a)=1\oplus 3 \\ g(a\ast c)=1\oplus 2 \\ g(c\ast a)=2\oplus 1 \\ g(b\ast c)=1\oplus 2 \\ g(c\ast b)=2\oplus 1 \\$

If all of these hold, then the map $g$ "preserves" the operation structures.

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    @Gomathi No, they did not ask a single question. It is labeled "example". They were simply *verifying* that the example truly fits their definitions. Stop trying to set the $a,b,c$ equal to any of the $1,2,3$ :)2012-08-23