# Proving that something is an isomorphism between group generators.

So I'm having trouble understanding isomorphisms between generators. In this problem, I had to show that a non-abelian group of order $$6$$ is isomorphic to $$S_3$$. Now I already showed that there were two elements of order $$2$$ and $$3$$, call them $$a$$ and $$b$$, and also proved that $$G=\langle a\rangle\langle b\rangle$$.

To show they are isomorphic, my initial thought was to construct a mapping defining the image of the generators $$a$$ and $$b$$ to generators of $$S_3$$ of the same order. For instance $$f(a)=(1,2)\text{ and } f(b)=(1,2,3).$$ Now I want to show that is an isomorphism.

First, the fact that it is a homomorphism is because of the way I constructed the map, defining the images of the generators, and then extended the images of the rest of the elements such that they satisfy the property of a homomorphism; that is, $$f$$ maps products and powers of generators to the products and powers of their images $$f(ab)=f(a)f(b)$$, and $$f(a^2)=f(a)^2$$, and because $$a$$ and $$b$$ are generators, then all the products and powers of them are the elements of $$G$$. So, by construction, this is a homomorphism.

Is this reasoning right?

(I'm starting with group theory so I may be lacking some accuracy in the proofs).

Then to show its an isomorphism, note that $$f(a^2=e)=e$$ and $$f(b^3=e)=e$$, because the generators map to elements (actually generators) of the same order in $$S_3$$. So the kernel is $$\{e\}$$ and it is injective.

To show it is surjective, I thought that it was because the generators of $$S_3$$ are the images of the generators of $$G$$ (again of the same order), so every element of $$S_3$$ has a pre-image. (Is this enough?). Therefore, because it is a homomorphism both injective and surjecive, then it is an isomorphism.

I would greatly appreciate any help or some clarity for my reasoning, thanks :)

• "isomorphism between generators" is senseless
– YCor
Mar 26 '19 at 7:46

In order to argue that this map is a homomorphism, you need to show that the generators of each of the groups satisfy the same relations. If you call the order $$2$$ generator of $$S_3$$ $$x$$ and the order $$3$$ generator of $$S_3$$ $$y$$, then $$S_3$$ can be described as the group generated by the elements $$x$$ and $$y$$ subject to the relations $$x^2 = e$$, $$y^3 = e$$, and $$yx = xy^{-1}$$.
If your $$a$$ and $$b$$ did not satisfy these same relations, i.e. if $$ba \neq ab^{-11}$$, then your map $$f$$ cannot be extended to a homomorphism (a fact that you should check).
Once you show that your $$a$$ and $$b$$ satisfy these same relations, then you can say without fear that you can extend the map on generators to a homomorphism.
To see why the proof is incomplete, let $$C_n$$ be the cyclic group of order $$n$$, so $$C_2\times C_3$$ is an abelian group generated by $$a,b$$, $$a^2=e=b^3$$. By mapping $$f(a)=(1,2)$$ and $$f(b)=(1,2,3)$$, your reasoning would give an isomorphism of $$C_2\times C_3$$ with a non-abelian group. Do go ahead and extend $$f$$ as you suggest and see where the "proof" breaks down.