# Homomorphism in algebraic systems

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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We'll need more information than this - how are the operations $*$ and $\oplus$ defined? – Matthew Pressland Aug 23 '12 at 15:20
@MattPressland I have edited the question. Please check it out. – Gomathi Aug 23 '12 at 15:53
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Aug 23 '12 at 16:01
@MartinSleziak Ok sir. Thank you. – Gomathi Aug 23 '12 at 16:05

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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Thanks. But how to find the right combination? Is that by trial and error method? – Gomathi Aug 23 '12 at 16:02
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. – Matthew Pressland Aug 23 '12 at 16:10
Ok sir. I get it. Thank you so much for your help. – Gomathi Aug 23 '12 at 16:12
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 $3$..." – Matthew Pressland Aug 23 '12 at 16:17
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 + ? – Gomathi Aug 23 '12 at 16:27

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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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. – Alexander Gruber Aug 23 '12 at 15:55
Do you mean we have to check for all three values (1,2,3) for a,b and c? – Gomathi Aug 23 '12 at 15:57
@AlexanderGruber Based on the OP's question, we have no reason to expect that there is an identity, or even associativity... – rschwieb Aug 23 '12 at 16:00
@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. – rschwieb Aug 23 '12 at 16:01
I have edited my question. Kindly check it out. – Gomathi Aug 23 '12 at 16:03