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I was studying binary operation on a set. Then the following question came to mind. I tried to find an answer. also searched in website but could not get any satisfactory answer.

the question is: is it possible to define two distinct binary operation on a same set?

the reason to search an answer for it is, if yes, then I think we can construct different group structure on a same set.

But thats the main problem. How to create a new binary operations from an existing one? I am searching for some recreational answer. please help me. In case it is already solved, kindly provide me the link

thanks in advance

KON3
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1 Answers1

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There are many groups which have the same cardinality. Suppose you have two groups $(G,\circ)$ and $(H,\cdot)$ and an arbitrary (set-theoretic) bijection $\varphi : G \to H$. You can define a new multiplication $\star$ in $G$ as follows : $$ g_1 \star g_2 \overset{def}= \varphi^{-1}( \varphi(g_1) \cdot \varphi(g_2)). $$ and $\star$ doesn't have to coincide with the multiplication in $G$, i.e. $\circ$. For instance, if $(G,\circ)$ and $(H,\cdot)$ are not isomorphic groups, then since $(G,\star)$ is isomorphic to $(H,\cdot)$ (the isomorphism being given by $\varphi : (G,\star) \to (H,\cdot)$), you see that $(G,\circ)$ and $(G,\star)$ are not isomorphic groups, so of course the binary operations $\star$ and $\circ$ have to be different.

Hope that helps,

Patrick Da Silva
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  • That was cool sir! very impressive and helpful idea for me. Thank you. Would you please specify what will happen for isomorphic case? – KON3 Aug 25 '14 at 10:08
  • And moreover sir, suppose that neither $G$ nor $H$ are groups but simply two ordinary sets. In that case, will your idea work? – KON3 Aug 25 '14 at 10:09
  • @AnjanDebnath : You can always make sure the operation you get from $H$ on $G$ is a different one by ensuring that your bijection maps the identity of $H$ to an element different from the identity of $G$.

    If $G$ nor $H$ are groups, I have no operation from $H$ to pullback on $G$ (which is essentially what I did). The idea is to take a group $H$ which you have, and for which $|G| = |H|$ (i.e. there exists a bijection $\varphi : G \to H$).

     – Patrick Da Silva Aug 25 '14 at 11:16
  • I use this bijection to take the group structure of $H$ and put it on $G$. If you pre-compose this bijection with a bijection from $G$ to itself, you can get different group structures on $G$, in particular by moving the identity element around.

    You can always start with a group $G$ and a bijection from $G$ to itself, and this will give you different (but isomorphic) group structures on $G$.

     – Patrick Da Silva Aug 25 '14 at 11:17
  • Patrick Da Silva, So, in case, if we have only one of $G$ and $H$ as group structure and the other as set, still it wont work? – KON3 Aug 25 '14 at 11:25
  • @AnjanDebnath This is a special case of transport of structure. – Bill Dubuque Aug 25 '14 at 12:15
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    @AnjanDebnath : If $H$ has a group structure then you can transport it to $G$, that is what I am doing. – Patrick Da Silva Aug 25 '14 at 13:05
  • Cool. Thank you so much – KON3 Aug 26 '14 at 04:45