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Let $G$ be a group and $H$ a subgroup of $G$. Is clear that if $x$ and $y$ are elements in $H$ then $[x,y] = x^{-1}y^{-1}xy \in H$. But, is true that, if $1 \neq [x,y] \in H$, then $x$ and $y$ are elements in $H$?

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2 Answers 2

up vote 3 down vote accepted

Suppose your claim holds. Let $x$ and $y$ be arbitrary non-commuting elements of an arbitrary group $G$. Now, $[x, y]\neq1$ will generate a cyclic subgroup of $G$, call it $H$. By your claim $x, y\in H$. Hence, $x$ and $y$ commute so $[x, y]=1$...a contradiction!

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Do you mean "will generate a cyclic subgroup"? –  Lee Mosher Aug 15 '14 at 14:53
@LeeMosher I may or may not have edited my answer after you posted your comment. But, in its current for, I am not sure what you mean... –  user1729 Aug 15 '14 at 14:54
$[x,y]$ itself is a single element, so I'm just pointing out that it doesn't quite make sense to say that it "forms" a cyclic subgroup. –  Lee Mosher Aug 15 '14 at 14:56
Great proof, by the way. –  Lee Mosher Aug 15 '14 at 14:56
@LeeMosher Ah, okay, thanks. And thanks for the complement - if you're ever looking for a postdoc... ;-) –  user1729 Aug 15 '14 at 14:57

It is false. Counterexamples comes from any non-abelian nilpotent group.

Consider for example the group of $3 \times 3$ upper triangular matrices with integer coefficients and $1$'s on the diagonal. If $$x = \pmatrix{1&1&0\\0&1&0\\0&0&1}, \quad y = \pmatrix{1&0&0\\0&1&1\\0&0&1} $$ then simple calculation shows that $[x,y]$ is an element of the subgroup $H$ consisting of all matrices of the form $$\pmatrix{1&0&k\\0&1&0\\0&0&1} $$

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