$G= \langle a, b : a^{7} = b^{3} = 1,\ b^{-1}ab = a^{2} \rangle$ and commutator group

Question

Consider the $a$ and $b$ the following permutation in $S_{7}$:

$$a = (1\ 2\ 3\ 4\ 5\ 6\ 7 ),\ b = (2\ 3\ 5)(4\ 7\ 6)$$

Consider the group $G = \langle a, b \rangle$. I know that $a^{7} = b^{3} = 1$ and $b^{-1}ab = a^{2}$. Moreover, with these relations we can say that $|G| = 21$.

Well, is simple to see that $\langle a \rangle \leq G'$, because

$$[a, b] = a^{-1}b^{-1}ab = a^{-1}a^{2} = a$$

I'd like to know that $G' = [G, G] = \langle a \rangle$. Someone can help me? Thank you.

Answer 1

To see the other inclusion, consider the map $G\rightarrow\mathbb{Z}_3$ sending $a$ to $0$ and sending $b$ to $1$. This application preserves relations, and so it induces an homomorphism from $G$ to $\mathbb{Z}_3$. Its kernel coincides with $\langle a\rangle$. Therefore $G/\langle a \rangle\cong\mathbb{Z}_3.$ By definition of commutator subgroup (as the smallest subgroup whose quotient is an abelian group), we obtain $G'=\langle a\rangle$.