Showing that two presentations are equivalent Are the groups $ \langle a,b;a^4,b^7,ab^3=ba^3 \rangle$ and $ \langle d;d^2 \rangle$ isomorphic?
I have been using the Tietze transformations to get from the first presentation into to the second, but I couldn't even show that it is commutative.
 A: I don't know anything about Tietze transformations, but the answer is yes. First, observe that the abelianization of the first group is $\mathbb{Z}_2$, generated by $a$. This is because the abelianization of the first presentation is
$$\langle a, b \mid a^4 = b^7 = 1, a^2 = b^2 \rangle$$
and from here we get $a^4 = b^4 = 1$, hence $b^{\gcd(4, 7)} = b = 1$, hence $a^2 = 1$. So there's a natural map from the first group to $\mathbb{Z}_2$, namely the abelianization map, and we'd like to show it's an isomorphism. 
This is just a matter of mucking around with the relations, and probably there are lots of ways to do it, but here's what I did. First, multiplying $ab^3 = ba^3$ by $a^3$ on the left and $a$ on the right we get
$$b^3 a = a^3 b.$$
We'll be using this "commutation" relation as well as the given one in what follows. Next, multiplying $ab^3 = ba^3$ by $b^4$ on the right and then manipulating it a bit using the "commutation" relations we get
$$a = b a^3 b^4 = b^4 a b^3 = b^5 a^3$$
and then multiplying by $b^2$ on the left and $a^3$ on the right we get
$$b^2 = a^2.$$
Now the argument proceeds as in the abelianization: this implies $b^4 = a^4 = 1$, hence $b = 1$, and hence $a^2 = 1$. 
