What is the order of a group given three generators, $a^2=b^3=c^4=1$ and $cb=ac$

Question

A group $G=\langle a,b,c\mid a^2=b^3=c^4=1, cbc^{-1} = a\rangle $ what is the order of the group $G$ give all such possible values.

My attempt: Since $cbc^{-1} =a \Rightarrow cb = ac\;\;(*)$, but then if multiply $(*)$ with $a$ from the left we get $$acb = c$$ And if we multiply $(*)$ with $b^2$ from the right we get $$c = acb^2$$ But then we get $acb^2 = acb \Rightarrow b^2 = b$ so $b=1$ then $a=1$. So the only options either $G = \langle c \rangle $ or $G=1$ the trivial group so either $|G|=4$ or $|G|=1$. Is that correct?

This is a question from a Preliminary algebra exam in a university.

Answer 1

By a Tietze transformation, the group $G$ given by the presentation in question is isomorphic to

$$\langle b,c\mid (cbc^{-1})^2=b^3=c^4=1\rangle.$$

But now $1=(cbc^{-1})^2=cb^2c^{-1}$, so that $1=c^{-1}c=c^{-1}(cb^2c^{-1})c=b^2$. Then $b=b(b^2)=b^3=1.$ This, again by Tietze transformation, means

$$G\cong\langle c\mid c^4\rangle.$$

The RHS is a presentation of $\Bbb Z_4$. Hence $|G|=4$.

Answer 2

Since $a$ and $b$ are conjugate, they have the same order. But the order of $a$ divides $2$, that of $b$ divides $3.$ Thus $a=b=e$.

Thus a presentation for $G$ is$\langle c\mid c^4\rangle, $ and we have $G\cong \Bbb Z_4.$