A unique operation on a set that makes it a group

Question

From Rotman's "Introduction to Group Theory":

Let $G$ be a group and let $X$ be a set having the same number of elements as $G$. If $f:G\rightarrow X$ is a bijection, there is a unique binary operation that can be defined on $X$ so that $X$ is a group and $f$ is an isomorphism.

Let $|G|=|X|=n$. Let the operation for $G$ be given by "$+$". Let us denote the operation for the set $X$ as "$\ast$" (we don't know if it's a group yet, though).

We can construct a bijection $f:G\rightarrow X$. Thus, $x=f(a)$ for every $x\in X$ and $a\in G$. We can see that $f(a+b)\in X\;\;\forall a,b\in G$.

My question is this: What should be my first step in approaching this problem? I was thinking of analyzing the equation:

$$f(a+b)=f(a)\ast f(b)$$ but can't think of anything useful to do with this.

I'm looking for hints on how to get started, not full answers please.

Thanks

$G$ is a group, $f$ is a bijection, hence there exists an inverse $f^{-1} \colon X \to G$ which is also a bijection. What can we say about the structure being preserved on $X$ by studying this inverse function to the group $G$? What happens if we assume that $f^{-1}$ is not a homomorphism? — Charles Boyd, Oct 28 '14 at 20:47
Consider that $$f^{-1}(x_{1}x_{2}) = f^{-1}(f(g_{1})f(g_{2})) = f^{-1}(f(g_{1}g_{2})).$$ — Charles Boyd, Oct 28 '14 at 20:54
This is an example of the ubiquitous technique of transport of structure. — Anne Bauval, Oct 10 '22 at 14:26

score 5 · Accepted Answer · answered Oct 28 '14 at 20:43

5

this is really just an exercise in abstraction. in essence we use the bijection to pull back the group structure on $G$ into the hitherto unstructured set $X$. we may represent the image of $x$ in $G$ as $x_G$ and if the inverse of this map is $\phi$ we may define an operation $\ast$ on $X \times X$ by: $$ x \ast y = \phi(x_G y_G) \\ x^{-1} = \phi((x_G)^{-1}) $$

answered Oct 28 '14 at 20:43

David Holden

18,040

So I think I understand what you are saying. Let me try to put it in my own words: can I simply define the operation $\ast$ to be $x\ast y = f( f^{-1}(x)+f^{-1}(y))$? Because then, for every $f(a),f(b)\in X$ (which correspond to $a,b\in G$), we would have: $f(a)\ast f(b) = f( f^{-1}[f(a)]+f^{-1}[f(b)])=f(a+b)$ for every $a,b\in G$, which satisfies homomorphism (and isomorphism, because of bijectivity). Is this what you meant? – Patrick Shambayati Oct 28 '14 at 23:56
Upon reviewing this, I am sure it is the right answer. I would be really surprised if it was not. $X$ seems to satisfy all the group criteria and the isomorphism is clear to me. Thanks. – Patrick Shambayati Oct 29 '14 at 00:05
1

yes. re your previous comment, if you want a fascinating example of this procedure, check out the so-called "wild" automorphisms of $\mathbb{C}$. it's to do with fields rather than just groups, but the principle is the same. it's an axiom-of-choice thing, but quite an eye-opener. there was a discussion of this on MSE http://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers – David Holden Oct 29 '14 at 00:45
if that is obscure you might first have a look at a similar but much simpler construction - Hamel bases for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ – David Holden Oct 29 '14 at 00:49

A unique operation on a set that makes it a group

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