2
$\begingroup$

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.

  • 5
    We'll need more information than this - how are the operations $*$ and $\oplus$ defined?2012-08-23
  • 0
    @MattPressland I have edited the question. Please check it out.2012-08-23
  • 0
    I've removed [tag:algebra] tag, since we don't use algebra tag anymore, see [meta](http://meta.math.stackexchange.com/questions/473/the-use-of-the-algebra-tag/3081#3081) for details.2012-08-23
  • 0
    @MartinSleziak Ok sir. Thank you.2012-08-23

2 Answers 2

1

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.

  • 0
    Thanks. But how to find the right combination? Is that by trial and error method?2012-08-23
  • 0
    Roughly, although you don't have to try every possibility. It's clear from the table that you have to either have $g(b)=1$ and $g(c)=2$ or $g(b)=2$ and $g(c)=3$, as $b,c$ and $2,3$ are the elements that give you the same answer no matter what you multiply by, and then checking how they multiply with $a$ and $1$ tells you which way round they have to be.2012-08-23
  • 0
    Ok sir. I get it. Thank you so much for your help.2012-08-23
  • 0
    That's good, because that comment has very confusing typos in it! And I can't edit it now. I meant "...$g(b)=2$ and $g(c)=1$, as $b,c$ and $1,2$...", and "...checking how they multiply with $a$ and $3$..."2012-08-23
  • 0
    Sir, I have another doubt. These two tables are of same size. But what if both are of different size. For eg. $+_4$ ([0],[1],[2],[3]) and B={0,1} with operation + ?2012-08-23
  • 0
    An isomorphism is in particular a bijection, so two algebraic structures with different cardinalities cannot be isomorphic.2012-08-23
  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/4603/discussion-between-gomathi-and-matt-pressland)2012-08-23
0

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.

  • 0
    May I add that it is usually a good idea to check whether the identity in $S$ maps to the identity in $P$ as the first step.2012-08-23
  • 0
    Do you mean we have to check for all three values (1,2,3) for a,b and c?2012-08-23
  • 0
    @AlexanderGruber Based on the OP's question, we have no reason to expect that there is an identity, or even associativity...2012-08-23
  • 0
    @Gomathi I don't know why you would do that. I am working under the assumption you had two different sets $\{1,2,3\}$ and $\{a,b,c\}$ which are in no way related to each other.2012-08-23
  • 0
    I have edited my question. Kindly check it out.2012-08-23
  • 0
    How to find the right values for a,b and c? Is that by trial and error method?2012-08-23
  • 0
    @Gomathi The page you linked to says to do exactly as I have just suggested to you. You have no reason to set things in either set equal to each other. The author clearly means that they are two sets of unrelated elements.2012-08-23
  • 0
    @Gomathi "Trial and error" is almost never a correct method in mathematics. You just have to *verify* everything I mentioned. "Trial and error" and "verify" are completely unrelated concepts.2012-08-23
  • 0
    Then, how did they find out the three values g(a)=3, g(b)=1 and g(c)=2 in first go itself?2012-08-23
  • 0
    @Gomathi Ah so it sounds like you're asking "how did they write this problem?" Well actually you can make up one of the multiplication tables randomly, then make up a bijection randomly, and then just write the second multiplication table accordingly. They wrote $a*b=a$, so that forces $g(a)\oplus g(b)$ to be $g(a)$. The rest of the second multiplication table would be *forced* upon us at this point.2012-08-23
  • 0
    So, where they asking the correct combination for which the problem will be isomorphic? I think I got confused a bit!2012-08-23
  • 0
    @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