everyone, how is everything going?
Can someone help me to identify the group $$\langle a,b| a^4=b^3=1, ba=a^3b \rangle$$
My guess is that this group can be identified bay $C_6$, but I am in trouble to show that it is abelian.
I know from the relations that \begin{align} a & =b^3a =b^2ba \\ &=b^2a^3b =b(ba)a^2b\\ & =ba^3ba^2b =(ba)a^2ba^2b \\ & = a^3ba^2ba^2b=a^3(ba)a(ba)ab \\ & = a^3a^3baa^3bab = a^2 b^2 ab \\ &= a^2b(ba)b = a^2ba^3bb \\ &= a^2 (ba)a^2bb = a^2a^3ba^2bb = aba^2bb.\end{align}
These calculations mean that $a=aba^2bb$, then $1=ba^2bb$, then $a^2=b^{-3}=1$. Hence, $ba=a^3b=a^2ab=ab$. Then this group is abelian.
Now if we define $G=\langle a,b \rangle$ such that $a,b$ satisfy those properties then we have $G=\{a^ib^j|i=1,2,3,4; j=1,2,3;ab=ba\}$ then $|G| \leqslant 6.$
By defining $h:\{a,b\} \to C_6$ such that $h(a)=r^3$ and $r(b)=r^2$ (seeing $C_6=\{1,r,r^2,r^3,r^4,r^5\}$) we can use the universal property of free groups to have an unique homomorphism $f:\langle a,b \rangle \to C_6$ such that $f(a)=r^3$ and $f(b)=r^2.$ Moreover $f$ is surjective since $\{r^3,r^2\}$ generates $C_6$.
Furthermore, $f(a^4)=f(a)^4=r^12=1$, $f(b^3)=f(b)^3=r^6=1$ and $f(bab^{-1}a^{-3}) = f(b)f(a)f(b)^{-1}f(a)^{-3}=r^2r^3r^{-2}r^{-9}=r^{-6}=1.$
Then $a^4, b^3 and bab^{-1}a^{-3}$ belongs to $Ker f$ and $\langle a,b \rangle / Ker f$ is isomorphic to $C_6$ by the first isomorphism theorem. This completes the proof that $C_6$ is isomorphic to our initial group.
I think the solution is "correct", but there are so many steps that I dont know for sure why they hold. I just follow the reasoning of my professor.
Can someone take a look and tell me what is not well explained in that proof?
Thank you very much!
You can milk the first part, which you did quite nicely, btw, for a lot more than you have.
You established that $a^2=1$, but this implies that $a^3=aa^2=a\cdot 1=a$. Thus your original presentation can be simplified to $<a,b|a^2=b^3=1,ab=ba>$. Now we are in a position to apply the fundamental theorem of finite abelian groups, which is a general theorem the describes the possible structures (up to isomorphism) for finite abelian groups. In the present case, this theorem say that there is only one finite group (up to isomorphism) generated by an order $2$ and an order $3$ element, namely, the cyclic group $C_6$. (see classification here: https://en.wikipedia.org/wiki/Abelian_group)