Isomorphism between two presentations of 6 order abelian group 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.
 A: 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.
A: 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.
