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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.

Community
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  • You don't know that $y \in C(G)$. If that was true then $G$ would be abelian! What you know is that $x^2 \in C(G)$ for all $x \in G$. So what does that tell you about the quotient group $G/C(G)$? – Derek Holt Nov 01 '15 at 16:59
  • @DerekHolt, yeahI just realized this. About to edit. –  Nov 01 '15 at 16:59

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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.

Nicky Hekster
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  • THANK YOU! Can you show me explicitly that every element in $G/N$ is order 2? I'd like to better be able to visualize it. Also, $x^{2}$ is normal b/c it's Abelian, right? –  Nov 01 '15 at 17:02
  • not to keep bugging you, but I am unfamiliar with the terminology $\gamma_{3}(G)$. Is it the same thing that I would call $G_{2}$, based on what I wrote above? –  Nov 01 '15 at 17:04
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    Every element of $G/N$ has order $2$ ($(xN)^2=x^2N=N$ for $x^2 \in N$) and it is well known and easy to prove that in general for a group in which every element has order $2$, then it is abelian. Now $N$ is normal since $g^{-1}x^2g=(g^{-1}xg)^2$, so conjugates of squares are again squares. – Nicky Hekster Nov 01 '15 at 17:06
  • In general, $\gamma_i(G)=[G,\gamma_{i-1}(G)]$ is the $i$-th term of the lower central series, a concept that is normally explained in connection to nilpotency. The lower central series reaches ${1}$ iff $G$ is nilpotent. See for example here https://en.wikipedia.org/wiki/Central_series – Nicky Hekster Nov 01 '15 at 17:11
  • I see. We were only told about the upper central series. –  Nov 01 '15 at 19:17
  • OK, different proof ($N$ as I defined), $G/Z(G)$ is abelian, since $G' \subseteq N \subseteq Z(G)$. Hence $Z(G/Z(G))=Z_2(G)/Z(G)=G/Z(G)$, so $G=Z_2(G)$, and you are done. – Nicky Hekster Nov 01 '15 at 23:19