Question

Prove that the group G=[a,b] with the defining set of relations a³=e, b⁷=e, a^-1ba=b⁸ is a cyclic group of order 3?

Answer 1

Consider the group $G=\langle a,b\rangle$ with the defining set of relations $a^3=e,\ b^7=e,\ a^{-1}ba=b^8.$ Since $b^7=e,$ we conclude that the equality $a^{-1}ba=b^8$ is equivalent to $a^{-1}ba=b^7b=eb=b,$ and hence (after left multiplication by $a$) to $ba=ab.$ It follows that the elements $a$ and $b$ commute. Since $3$ and $7$ are relatively prime, we conclude that the element $ab$ is of order 21, that is $(ab)^{21}=e.$ Therefore, $G$ is a cyclic group of order 21.