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Assume $x,\,y\in G$ and both commute with $[x,\,y]$. Prove that for all $n\in\mathbb{Z}^+,\;(xy)^n=x^ny^n[y,\,x]^{\frac{n(n-1)}{2}}$.

$[x,\,y]=x^{-1}y^{-1}xy$ is the commutator of $x$ and $y$.

I have found that \begin{equation} xy^{-1}xy=y^{-1}xyx\\ yx^{-1}y^{-1}x=x^{-1}y^{-1}xy\\ [x,\,y][y,\,x]=1\\ x=[y,\,x]x[x,\,y]\\ y=[y,\,x]y[x,\,y]. \end{equation} However, I can only prove the special case when $n=2$.

user26857
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Knt
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    Hint: Use Induction – Kumar Jul 09 '19 at 04:22
  • I have thought of that. Can you give more details? – Knt Jul 09 '19 at 04:23
  • You need to use these three things with induction: $1:$ $xy=[y,x]xy[x,y]$; $2:$ $[x,y][y,x]=[y,x][x,y]=1$ ;$3:$ $x,y$ don't commute with each other but they commute with $[x,y]$ as well as $[y,x]$ – Kumar Jul 09 '19 at 04:39
  • You can also check out this Question: https://math.stackexchange.com/questions/282590/check-in-detail-that-langle-x-y-mid-xyx-1y-1-rangle-is-a-presentation?rq=1 – Kumar Jul 09 '19 at 04:42

1 Answers1

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Note that if $ab=ba$, then $ab^{-1}=b^{-1}a$. Applying this result to $x,y$ and $[x,y]$, we get that $[y,x]$ also commute with $x$.

For $n=2$, $$ xyxy=xxy[y,x]y=xxyy[y,x]=x^2y^2[y,x], $$ as required.

Now suppose that $(xy)^n=x^ny^n[y,\,x]^{\frac{n(n-1)}{2}}$, then $$ (xy)^{n+1}=xy(xy)^n=xyx^ny^n[y,\,x]^{\frac{n(n-1)}{2}}\\ =xxy[y,x]x^{n-1}y^n[y,\,x]^{\frac{n(n-1)}{2}}\\ =x^2yx^{n-1}y^n[y,\,x]^{\frac{n(n-1)}{2}+1}. $$ Now use the fact that $[x,[x,y]]=[y,[x,y]]=e$ repeatedly, $$ x^2yx^{n-1}y^n[y,\,x]^{\frac{n(n-1)}{2}+1}\\ =x^{1+k}yx^{n-k}y^n[y,\,x]^{\frac{n(n-1)}{2}+k}. $$ Taking $n=k$, we obtain $$ (xy)^{n+1}=x^{n+1}y^{n+1}[y,\,x]^{\frac{n(n+1)}{2}}. $$ By the induction principle, we finish the proof.

Junglemath
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Ma Joad
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