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From this link we can see the order $6$ abelian group can have two presentations:

  • One generator: $\langle k\mid k^6\rangle $

  • Two generators: $\langle k,r\mid k^3, r^2, krk^{-1}r\rangle $

Both of them are generating the cyclic group of order six.

We know there's a unique abelian group of order six, i.e. the cyclic one, up to isomorphisms.

My aim would be to find this isomorphism explicitly.

I thank you for any help in advance.

Shaun
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riccardoventrella
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2 Answers2

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I think to have found it explicitly.

Let's redefine the two groups like this:

$$A=\langle r\mid r^6\rangle \\ B=\langle a, b\mid a^3, b^2, aba^{-1}b\rangle $$

We can define a homomorphism $\Phi = A\mapsto B$, where $\Phi(xy)=\Phi(x)\Phi(y), \forall x, y\in A$, like that:

$$\Phi(r^2)=a, \Phi(r^3)=b$$

Let's check this is a legitimate isomorphism. Let's expand the $A$ and $B$ groups elements:

$$A=\{e_A, r, r^2, r^3, r^4, r^5\} \\ B=\{e_B, a, b, ab, a^2, a^2b\}$$

We also known from $B$ relator that $ab=ba$. Going through each element, we get:

$$\Phi(r^2)=a \\ \Phi(r^3)=b \\ \Phi(r^4)=\Phi(r^2r^2)=\Phi(r^2)\Phi(r^2)=a^2 \\ \Phi(e_A)=\Phi(r^2r^4)=\Phi(r^2)\Phi(r^4)=a^3=e_B \\ \Phi(r)=\Phi(r^3r^4)=\Phi(r^3)\Phi(r^4)=a^2b \\ \Phi(r^5)=\Phi(r^2r^3)=\Phi(r^2)\Phi(r^3)=ab$$

Let's check the element order is preserved by the mapping, pointwise.

$$|r^2|=3, |a|=3 \\ |r^3|=2, |b|=2 \\ |e_A|=2, |e_B|=2 \\ |r^4|=3, |a^2|=3 \\ |r|=6, |a^2b|=6, since (a^2b)^2=a^2baab=a^2abab=bab=ab^2=a \implies (a^2b)^6=a^3=e_B \\ |r^5|=6, |ab|=6, since (ab)^6=a^6b^6=(a^3)^2(b^2)^3=e_B$$

So $\Phi$ maps the identity of A to the idendity of B, it's one to one and onto B, preserving all elements order pointwise.

So $\Phi$ it's an isomorphism.

riccardoventrella
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    The relator in the second group should be $aba^{-1}b^{-1}$. Otherwise what you have is probably $S_3\cong \Bbb Z_3\rtimes \Bbb Z_2$. Oh I see, $b=b^{-1}$. Nevermind. – calc ll Aug 14 '22 at 11:22
  • @Noobie, Yes $b=b^{-1}$ since $b^2=e$. However $S_3$ would have had $abab$ as relator – riccardoventrella Aug 14 '22 at 14:07
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One can use von Dyck's Theorem to obtain the morphisms; this will save you the work of checking the maps are homomorphisms.

Let $$\begin{align} G&= \langle k\mid k^6\rangle,\\ K&= \langle s,r\mid s^3, r^2, srs^{-1}r\rangle. \end{align}$$ (I changed the name of the first generator in the second group so we have different names in the two groups, to avoid potential confusion.)

To define a morphism $G\to K$, we need to find an element in $K$ of exponent $6$. To define a morphism $K\to G$, we need to find two elements in $G$ that satisfy the relations $r$ and $s$ do in $K$.

For the latter, consider $k^2$ and $k^3$. We have $(k^2)^3=k^6=1$, $(k^3)^2=k^6=1$, $(k^2)(k^3)(k^2)^{-1}(k^3)=k^{2+3-2+3}=k^6=1$. So we have a group homomorphism $\phi\colon K\to G$ given by $\phi(s)=k^2$, $\phi(r)=k^3$.

In the other direction, note that $s$ and $r$ commute, as $1= srs^{-1}r=srs^{-1}r^{-1}=(sr)(rs)^{-1}$ (since $r^2=1$ implies $r=r^{-1}$). So $sr=rs$.

Now consider $s^{-1}r$. We have $(s^{-1}r)^6 = s^{-6}r^6 =(s^3)^{-2}(r^2)^3=1$. So we have a morphism $\psi\colon G\to K$ given by $\psi(k)=s^{-1}r$.

And they are inverses of each other: $$\begin{align} \phi(\psi(k))&= \phi(s^{-1}r) = \phi(s)^{-1}\phi(r)=k^{-2}k^3=k;\\ \psi(\phi(s))&= \psi(k^2)=(s^{-1}r)^2= s^{-2}r^2=s,\\ \psi(\phi(r))&= \psi(k^3)=(s^{-1}r)^3= s^{-3}r^3= r. \end{align}$$.

So you have your explicit isomorphisms.

Arturo Magidin
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  • Thanks @Arturo. So the map I've defined Is equivalent to yours, right? – riccardoventrella Aug 14 '22 at 22:15
  • I mean, my $\Phi$ should be enugh tio define an isomorphism, since one to One and onto. – riccardoventrella Aug 14 '22 at 22:21
  • @riccardoventrella Yes. You have $\Phi(r)=\Phi(r^3r^{-2})=ba^{-1}$ which is the same as my $\psi$, though I would say you give a convoluted definition, as you never say explicitly what happens to the generator. You are "really" defining the inverse map but presenting it as going the other way. – Arturo Magidin Aug 14 '22 at 22:24
  • @riccardoventrella: Here's the problem: you tell us what $\Phi(r^3)$ is; you tell us what $\Phi(r^2)$ is. Unless you assume $\Phi$ is multiplicative, this doesn't tell us what $\Phi(r)$ is. And of course, you can't just assume that $\Phi$ is multiplicative, and then tell us that you can prove it is multiplicative. If you are going to define it as a function and then check that it is a group homomorphism, then you need to tell us exactly how you define it at every element of the domain, which you did not really do. – Arturo Magidin Aug 15 '22 at 00:08
  • @riccardoventrella This is precisely the issue that von Dyck's theorem avoids: the theorem ensures that what we have is a group homomorphism, so we don't need to check whether it is multiplicative, and it is completely determined by what it does on the generators. – Arturo Magidin Aug 15 '22 at 00:37
  • I saw morphisms defined pointwise like I did in many textbooks and alsonin many automorphism related answers on this site Of course your definition Is more technically complete and confirms my intuition about It. – riccardoventrella Aug 15 '22 at 06:51
  • However I will define better my answer showing how It Will act on all elements even if I thought It was understood. – riccardoventrella Aug 15 '22 at 07:12