# Group Presentation, Identification

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!

• Note that, since you have shown that $a^2=1$ it follows that $|a|=2$ (not $4$), therefore, $ba=a^3 b$ becomes $ba=ab$ (why?), thus we have that $a^2$ and $b$ in fact commute. Two coprime order elements that commute generate a cyclic (sub)-group. So it is indeed $C_6$. – Justin Benfield Oct 23 '16 at 17:06
• Broadly speaking, this proof has 3 steps. The section from "I know from the relations..." to "...$|G|\leq 6$" is converting the presenation given into a more sensible one $\langle a,b|a^2=b^3=1, ab=ba \rangle$. The next paragraph defines a homomorphism, and the remainder shows that it is in fact an isomorphism. – Eric Stucky Oct 23 '16 at 17:06
• Thank you @JustinBenfield I have never thought this way! – bttmbrcelo Oct 23 '16 at 17:12
• Thank you @EricStucky for your feedback. Do you think the proof is ok? I am not sure of why I can do that things in the final, because I know that $\langle a, b \rangle / ker f$ is isomorphic to $C_6$, but why is it implies that my set is in fact isomorphic to $C_6$? – bttmbrcelo Oct 23 '16 at 17:12
• (Yikes, sorry, my explanation of step 3 is wrong: the homomorphism is not an isomorphism, but rather it defines an isomorphism on a quotient.) The last paragraph worries me a little bit: you've shown me elements in the kernel, but how do you know that $\ker f$ isn't larger than the subgroup they generate? [Haha, same issue: yes, you need to show that $\ker f$ is exactly equal to $(a^4, b^3, bab^{-1}a^{-3})$. For this it would probably help to use the new presentation from the first step, which you haven't really used yet.] – Eric Stucky Oct 23 '16 at 17:20

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)