The group generated by a and b is the same as the group generated by ab and b

Question

For a group G generated by two elements a and b, both in G with order 2, show that the group generated by a & b is the same as the group generated by ab & b.

Would I have to somehow show they're identical through finding the identities and comparing?

Answer 1

Actually, this statement holds in full generality.

Then, you can express both $ab$ and $b$ using the building blocks $a$ and $b$ (and their inverses).
This implies the containment of the generated subgroups $\langle ab,b\rangle\subseteq \langle a,b\rangle$.

And conversely, you can express both $a$ and $b$ using the building blocks $ab$ and $b$ (and their inverses), so that $\langle a,b\rangle\subseteq \langle ab,b\rangle$.

We have $a=(ab)\cdot b^{-1}$.