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Let $G$ be a group and $U=\{xyx^{-1}y^{-1}:x,y\in G\}.$ How to show that for $g\in G,u\in U,~gug^{-1}\in U?$

Added: Actually I was trying to show that the commutator subgroup of $G$ is normal in $G$ and I've in hand the result which says for $U\subset G,gug^{-1}\in U~\forall~g\in G,u\in U\implies(U)\lhd G.$

Sriti Mallick
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  • Have you tried anything so far? Given a $u\in U$, how else might you write it? And once you have it in this form, does the conjugate $gug^{-1}$ look similar to elements of $U$? How might you alter your expression to make it look like an element of $U$? – Bey Apr 11 '13 at 12:31
  • Note: $U$ is not in general a subgroup of $G$. – user1729 Apr 11 '13 at 13:52

1 Answers1

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$$gxyx^{-1}y^{-1}g^{-1}=(gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1}$$

DonAntonio
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