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Derek Holt
Derek Holt
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From the way you written it, it looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = |g^n|_S/|n|$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.

Let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

Another familiar example is the Heisenberg group $\langle x,y \mid [[x,y],x]=[[x,y],y]=1 \rangle$, with $S=\{x,y\}$. Then $|x^n|_S = |y^n|_S=n$, but $[x,y]^{n^2}=[x^n,y^n]$, so $|[x,y]^{n^2}|_S \le 4n$.

From the way you written it, it looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = |g^n|_S/|n|$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.

Let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

From the way you written it, it looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = |g^n|_S/|n|$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.

Let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

Another familiar example is the Heisenberg group $\langle x,y \mid [[x,y],x]=[[x,y],y]=1 \rangle$, with $S=\{x,y\}$. Then $|x^n|_S = |y^n|_S=n$, but $[x,y]^{n^2}=[x^n,y^n]$, so $|[x,y]^{n^2}|_S \le 4n$.

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Derek Holt
Derek Holt
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For a specific exampleFrom the way you written it, letit looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = |g^n|_S/|n|$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.

Let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

For a specific example, let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

From the way you written it, it looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = |g^n|_S/|n|$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.

Let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.

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Derek Holt
Derek Holt
  • 37.4k
  • 4
  • 95
  • 149

For a specific example, let $G$ be the Baumslag-Solitar group ${\rm BS}(1,2) = \langle x,y \mid y^{-1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.

Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $|y^n|_S = |(xy)^n|_S = n$, but $x^{2^n} = y^{-n}(xy)(y^{-1})y^n$, so $|x^{2^n}|_S \le 2n$.