Why might $\langle a^2, b\rangle$ be unique? Why might the subgroup $H=\langle a^2, b \rangle \triangleleft \langle a, b \mid a^4 = b^{17} = 1, aba^{-1} = b^{-1}\rangle = G$ be the unique subgroup of index $2$?
I've tried some arguments I've seen on here; if $K$ is another such subgroup then I can show that $\langle a^2, b\rangle \cap K$ has order $17$, but I'm not sure how to proceed to a contradiction from this fact (or if one can).
Also, since $\langle a^2, b \rangle$ is normal, it's the kernel of a homomorphism $\phi : G \to \Bbb Z/(2)$, so presumably I might be able to show that there is only one such homomorphism? Otherwise I'm not entirely sure how I might prove this. Unless it's wrong or entirely trivial and I've missed the point!

Good job figuring out that the subgroup $A$ generated by $b$ must be $\le K$.
If you have covered the correspondence theorem then you can argue as follows (justify the claims). 


*

*The subgroup $A=\langle b\rangle\unlhd G$.

*$G/A\simeq C_4$.

*By correspondence theorem the subgroups of $G$ containing $A$ are in bijective correspondence with subgroups of $G/A$.

*$G/A$ has only a single subgroup of index two.

