This is related to a question I asked yesterday.
Let $G$ be a group such that for every $x$, $y$ $\in G$, the elements $x^{2}$ and $y$ commute. I need to show that $G$ is nilpotent.
This is what I have so far:
Since $\forall x,\,y \in G$, the elements $x^{2}$ and $y$ commute, $\{x^{2}\} \in C(G)$, the center of $G$.
Now, $G_{0} = G$, $G_{1}=G/C(G)=\{\text{elements of the form}\,zC(G),\,\text{where}z\in G\}$.
But, 1) I'm not sure if $x^{2}$ is even all of $C(G)$ (there could be other elements in $G$ that also commute) or even how to show that it is if in fact it is. Also, 2) If I have it done right so far, how do I figure out what $G_{2}$ is, and whether or not it is trivial?
I say that this problem is related to the other one I asked about because showing that a group is nilpotent or solvable both kind of deal with looking at quotient groups/cosets, and I still don't entirely understand those.
Could somebody help me with this, and in the process, help me to understand? Pretend like you're trying to explain this to somebody who is incredibly clueless, because that is exactly how I feel right now.
Consider the subgroup $N=\langle x^2 : x \in G \rangle$. Then $N$ is central and $G/N$ is abelian, since every element has order $2$ in this quotient. Hence $G' \subseteq N$, so $\gamma_3(G)=[G',G]=1$ and $G$ is nilpotent.