Prove that $\langle a,b\mid ab=ba^n\rangle$ is not isomorphic to $\langle a,b\mid ab=ba^{n'}\rangle$ when $n\neq n'$

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For a positive integer $n$, let $X_n$ be be the quotient space $\left([0,1] \times S^1\right) / \sim$, where the equivalence relation $\sim$ is generated by $$ (0, z) \sim\left(1, z^n\right), \quad \forall z \in S^1 . $$ (a) Compute $\pi_1\left(X_n\right)$ in terms of generators and relations.

(b) For any different $n, n^{\prime}>0$, show that $X_n$ is not homotopy equivalent to $X_{n^{\prime}}$. $$$$ By using CW-complex we can verify that $\pi_1\left(X_n\right)=\langle a,b\vert ab=ba^n\rangle$, and to prove (b) we only need to prove that $\langle a,b\mid ab=ba^n\rangle$ is not isomorphic to $\langle a,b\mid ab=ba^{n'}\rangle$ when $n\neq n'$, but I don't know how to prove that two groups are not isomorphic. Are there any general ways (that can be used here) to prove two groups are not isomorphic? Or should I choose other potential ways here? Thanks in advance!

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The abelianization of $\langle a,b\mid ab=ba^n\rangle$ is isomorphic to $\Bbb Z\times\Bbb Z/(n-1)\Bbb Z,$ whose torsion subgroup is of order $n-1.$