How do I find the derived subgroup for this particular group?

Question

In describing the character table for groups in the Representation and Characters of Groups by James, I came upon this particular group $G=\langle a,b: a^{2n} = b^4 = 1, ba = a^{-1}b^{-1},b^{-1}a=a^{-1}b \rangle$ of order $8n$. This book has a solution section, which states as a matter of fact that the derived subgroup $G'$ is $\langle a^2, b^2 \rangle$. However, I find no easy way to do so. I did come up with the relation that is $a^ib^j = b^{j(-1)^i}a^{i(-1)^j}$ for every $i,j \in \mathbb Z$. But trying to compute $[a^{i_1}b^{j_1},a^{i_2}b^{j_2}]$ produces a very long winded expression which might work, but is inefficient and inelegant. That is why I'm making this thread, seeking a better solution.

Answer 1

If you are given a presentation, it's easy to understand the quotient by the derived subgroup: pretend the all of the generators commute, i.e., add $ab=ba$ as a relation. The presentation then becomes $$ \langle a,b\mid a^{2n}=b^4=1,\,a^2b^2=1,\,a^2b^{-2}=1,(ab=ba)\rangle.$$ Any word that resolves to the trivial word in this group is in the derived subgroup.

If $n$ is even then this is isn't $C_2\times C_2$. If $n$ is odd, then we see that $a^2=b^2$, and $b$ has order dividing $4$. Thus $a$ actually must have order dividing both $2n$ and $4$, i.e., $2$. Then $a^2=b^2$ shows that $b$ has order $2$, and both $a^2$ and $b^2$ lie in $G'$.