# Claiming a finite set $S$, on which there is a binary operation $\bot$, to be a group by examining its multiplication table

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.

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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.

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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. – Dan Rust Jan 21 '13 at 21:51

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).

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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