I want to check if my solution to one problem from my group theory course is valid. The problem is:
Given $D_n=\{x^iy^j:0\leq i<n,0\leq j<2\}$, prove that $D'_n=\langle x^2\rangle$.
My attempt is:
Given two different elements $g=x^ay^b$ and $h=x^cy^d$ (it's clear that $g,h\in D_n$). We calculate its commutator:
$$[g,h]=x^ay^bx^cy^dy^{-b}x^{-a}y^{-d}x^{-c}$$ Using that $yx^iy=x^{-i}$ (I'll go almost straight to the result avoiding the operations) we get that
$$[g,h]=x^ax^{-c}y^by^dy^{-b}y^{-d}x^ax^{-c}=x^{2a-2c}=(x^2)^{a-c}\in \langle x^2\rangle.$$ So this proves that, $\forall g,h\in D_n$, $[g,h]\in\langle x^2\rangle$, so from this we conclude that $$\boxed{D'_n=\langle x^2\rangle}$$ Is my solution valid? If not, why? Any help will be appreciated, thanks in advance.
Well, you showed that $D_n'\subseteq \langle x^2\rangle$. But what about the other direction? It is easy though, because $x^2=[x,y]\in D_n'$, and so $\langle x^2\rangle\subseteq D_n'$.
By the way, another way to see that $D_n'\subseteq\langle x^2\rangle$ is to note that $D_n/\langle x^2\rangle$ has either order $2$ or order $4$, depends if $n$ is even or odd. But in either case, $D_n/\langle x^2\rangle$ is abelian, and so $\langle x^2\rangle$ contains all the commutators of the group.