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Stefan Kohl ♦
Stefan Kohl ♦
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Can group solvability be dected bydetected from identities among the generators?

For "$n=1$" the answer is "yes." A-- A group is abelian iff its generators commute.

Let $G_0=G$ be a group and let it be generated by $X_0=X$. For each $n>0$ let $G_n=[G_{n-1},G_{n-1}]$ and let $X_n=[X_{n-1},X_{n-1}]$. If $X_n$ is trivial, is $G_n$ trivial?

Can group solvability be dected by identities among the generators

For "$n=1$" the answer is "yes." A group is abelian iff its generators commute.

Let $G_0=G$ be a group and let it be generated by $X_0=X$. For each $n>0$ let $G_n=[G_{n-1},G_{n-1}]$ and let $X_n=[X_{n-1},X_{n-1}]$. If $X_n$ is trivial, is $G_n$ trivial?

Can group solvability be detected from identities among the generators?

For $n=1$ the answer is "yes." -- A group is abelian iff its generators commute.

Let $G_0=G$ be a group and let it be generated by $X_0=X$. For each $n>0$ let $G_n=[G_{n-1},G_{n-1}]$ and let $X_n=[X_{n-1},X_{n-1}]$. If $X_n$ is trivial, is $G_n$ trivial?

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Matt Brin
Matt Brin
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Can group solvability be dected by identities among the generators

For "$n=1$" the answer is "yes." A group is abelian iff its generators commute.

Let $G_0=G$ be a group and let it be generated by $X_0=X$. For each $n>0$ let $G_n=[G_{n-1},G_{n-1}]$ and let $X_n=[X_{n-1},X_{n-1}]$. If $X_n$ is trivial, is $G_n$ trivial?