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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 :)

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    $\begingroup$ "isomorphism between generators" is senseless $\endgroup$
    – YCor
    Mar 26 '19 at 7:46
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This proof as written is not entirely correct.

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.

A small note about proving bijectivity: it's an easy fact (proved by just counting elements) from set theory that if you have an injection (or a surjection) between finite sets of the same size, then this injection (or surjection) must in fact be a bijection. So once you've proved that f is either an injection or a surjection, it follows that it must be a bijection as well. In your case it's relatively easy to see both injectivity and surjectivity, but in more complicated cases it can be useful to know that for finite groups of the same size it suffices to show only one of the 2.

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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.

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