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If $H \leq G$ is generated by elements of the form $xyx^{-1}y^{-1}$, does this mean that one can define the group $H$ as $<xyx^{-1}y^{-1}> = \{(xyx^{-1}y^{-1})^n: x, y \in G, n \geq 1\}$?

sequence
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    Remember it would also include things of the form $(xyx^{-1}y^{-1})(wzw^{-1}z^{-1})$, where $w,z$ are different than $x$ and $y$. – Ben Sheller Oct 11 '15 at 20:09
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    I thought it is generated by elements of a certain form. Doesn't your notation seem to imply it's generated by a single element? – pjs36 Oct 11 '15 at 20:09
  • No, it does not mean that. It does not even imply that, but it is not easy to give an example. – Mariano Suárez-Álvarez Oct 11 '15 at 20:09
  • @MarianoSuárez-Alvarez Supplementary question: What is a minimal counterexample (to pjs36's suggestion)? – Travis Willse Oct 11 '15 at 20:12
  • Moreover, you wrote «if $H$ is generated by elements of the form...»: do you mean that it is generated by some elements of that form or by all elements of that form? – Mariano Suárez-Álvarez Oct 11 '15 at 20:15
  • @Travis, one could precise the question into «What is the minimal size of a group whose derived subgroup has elements which are not powers of commutors?» I have no idea :-) – Mariano Suárez-Álvarez Oct 11 '15 at 20:17
  • @Travis Is this the sort of thing you're asking about? My suggestion was merely that OP's notation is not what they mean to be writing. – pjs36 Oct 11 '15 at 20:19
  • @MarianoSuárez-Alvarez, it's not specified in the problem. But what is said is that "$H$ is called the commutator subgroup of $G$". – sequence Oct 11 '15 at 20:32
  • @pjs36 Thanks! (I had just found that post and was about to link it myself!) As you probably saw, this post is about a slightly more restrictive question, and asks for a group with an element that is not a commutator---rather than not some power of a commutator. I do not know whether the examples in the answer are powers of commutators, but in either case, this means the phenomenon does not occur in groups of order $< 96$. – Travis Willse Oct 11 '15 at 20:41
  • (I only noticed now that the examples in the answer I referenced were given by @MarianoSuárez-Alvarez himself.) – Travis Willse Oct 11 '15 at 20:46

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@sequence - the notation should be $G'=\langle x^{-1}y^{-1}xy: x, y \in G \rangle$. So do not forget to let vary $x$ and $y$ over the whole of $G$. If you write $\langle x^{-1}y^{-1}xy\rangle$, then it is the (cyclic) group generated by $x^{-1}y^{-1}xy$.

Nicky Hekster
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  • The OP wrote, at the very end of the current form of the question, "$={(xyx^{-1}y^{-1})^n:x,y\in G, n\geq 1}$, so it seems the incorrect notation was intentional --- except that the OP wrote the semigroup rather than the group generated by a single commutator. So the short answer to the question is "no". – Andreas Blass Oct 11 '15 at 22:12
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No, it means that $H$ is the subgroup generated by all the commutators $xyx^{-1}y^{-1}$, not just by a single one. So $H$ consists of all products of (possibly different) commutators, not just powers of a single one.

Ordinarily, "subgroup generated" would also require you to include the inverses of the generators. In the present special situation, that's superfluous, because the inverse of a commutator $xyx^{-1}y^{-1}$ is another commutator, $yxy^{-1}x^{-1}$. But in the question, where you used powers of a single commutator, what you wrote would not even be a group in general because, by writing $n\geq1$, you didn't put inverses or the identity element into your $H$.

Andreas Blass
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