Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be a finite set on which there is a binary operation $\bot$ defined by a given multiplication table. Is it mathematically correct to claim that $(S,\bot)$ is a group because the multiplication table of $\bot$ matches the multiplication table of a known finite group $(G, *)$?

I guess one can say that this is obviously true, but I am interested in a formal argument.

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

This is true because you can define an isomorphism between the elements of $S$ and $G$ in the obvious way. That is, a bijective function $f\colon S\rightarrow G$ such that $f(a)f(b)=f(a\bot b)$. If $S$ with the defined binary operation is isomorphic to a group, then you should easily be able to show that all the group axioms hold on $S$. As a quick example I'll show that all $a\in S$ have a right inverse.

Let $c=(f(a))^{-1}$. $f$ is a bijection and so there exists a $b\in S$ such that $f(b)=c$. Now, $f(a)(f(a))^{-1}=1$ which means that $f(a)c=1$. $f$ is an isomorphism and so $f^{-1}$ is also an isomorphism, hence $f^{-1}(f(a))\bot f^{-1}(c)=f^{-1}(f(a).c)=f^{-1}(1)$ which means that $a\bot b=1$ and so $a$ has a right inverse.

You can show the rest of the axioms hold.

share|improve this answer
    
Thank you Daniel for your detailed answer. The comment I made to spin's answer applies to yours as well. –  gpo Jan 21 '13 at 21:42
    
You only need to use the fact that $f$ is a bijection with the property that $f(a)f(b)=f(a\bot b)$. I perhaps called it an isomorphism pre-emptively but you can ignore that. You can also prove directly that the inverse of such a function has the same structure preserving property (you might like to prove this yourself) and so there's no trouble with the example I gave for existence of inverses, or any of the other axioms for a group. –  Daniel Rust Jan 21 '13 at 21:51
add comment

Yes, this is true.

If $(S, \bot)$ and $(G, *)$ have the "same multiplication table", then there is a bijection $f: G \rightarrow S$ such that $f(x * y) = f(x) \bot f(y)$ for all $x, y \in G$. From this it follows that $(S, \bot)$ is a group isomorphic with $(G, *)$ (you might want to prove this).

share|improve this answer
1  
That was my initial thought too. But the objective here is to prove that $(S,\bot)$ is a group. The definition of a group isomorphism involves two groups. So, to my understanding, one must establish that $(S, \bot)$ is a group before trying to prove that there is an isomorphism between $(S, \bot)$ and $(G, *)$. No? –  gpo Jan 21 '13 at 21:40
    
Yes, what I was trying to say is that you should use the bijection $f$ to show that $(S, \bot)$ is a group. After that the fact that $f$ is a group isomorphism is immediate. –  spin Jan 21 '13 at 21:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.