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I guess that multiplication of commutators isn't equal to commutator of some elements. But I don't know how to find such group. Because my task became so : I tried to get four elements (for example in $GL_{2}$) , consider their product of commutators and then I stuck because it become difficult to prove that there is no such elements $f$ and $g$ : product = $[f,g]$. Any ideas?

William Elliot
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openspace
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    If you search this site you will find examples of groups in which the derived subgroup contains elements which are not commutators. In a finite example of that there exists a pair of commutators whose product is not a commutator. – Mariano Suárez-Álvarez Oct 11 '17 at 12:01
  • The set of commutators is closed under inverses, so you will find an example in any group where the set of commutators is not equal to the derived subgroup of $G$. Some examples are given here. If you want an explicit example of two commutators such that their product is not a commutator, you have to do a little bit more work. – Mikko Korhonen Oct 11 '17 at 13:38

2 Answers2

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In general, $[G,G] \not = \{[g,h] :g,h \in G\}$. We shall illustrate this via $\operatorname{SL}(2,\mathbb{R})$, the group of real $2 \times 2$ matrices with determinant 1.

First we show $-I$ is not a commutator. Suppose, to the contrary, that $A^{-1}B^{-1}AB = -I$ for some $A,B \in SL(2,\mathbb{R})$. Then considering traces yields, $$ \operatorname{tr}(A) =\operatorname{tr} (B^{-1}AB) = \operatorname{tr}(-A)$$ And therefore, $$ A=\left( \begin{array}{cc} a_{11}&a_{12} \\ a_{21}&-a_{11}\\ \end{array} \right) \quad \text{for some} \quad a_{11},a_{12},a_{21} \in \mathbb{R} .$$ Since $\det(A)= -{a_{11}}^2 -a_{12}a_{21} =1$ we have that $a_{12} \neq 0 \neq a_{21}$ and $A^2 =-I$. That is $\left( \begin{array}{c} 1 \\ 0\\ \end{array} \right)$ and $A \left( \begin{array}{c} 1 \\ 0\\ \end{array} \right)$ are linearly independen; with respect to this basis, $ A=\left( \begin{array}{cc} 0&-1 \\ 1&0\\ \end{array} \right) .$ Thus, $$ -\left( \begin{array}{cc} b_{11}&b_{12} \\ b_{21}&b_{22}\\ \end{array} \right)= -B = ABA^{-1} = \left( \begin{array}{cc} b_{22}&-b_{21}\\ -b_{12}&b_{11} \\\end{array} \right), $$ and wee see that $b_{11} = b_{22}$ and $b_{12} = b_{21}$. So $\det(B) = -{ b_{11}}^2 -{ b_{12}}^2 \leq 0$, contrary to $B \in \operatorname{SL}(2,\mathbb{R})$. A contradiction!

It remains to show that $-I \in \left[ \operatorname{SL}(2,\mathbb{R}), \operatorname{SL}(2,\mathbb{R}) \right]$. To this end verify that, \begin{align*} X&= \left( \begin{array}{cc} 1&0 \\ 1&1\\ \end{array} \right) =\left[ \left( \begin{array}{cc} 1&-1 \\0&1\\ \end{array} \right) , \left( \begin{array}{cc} \sqrt{2}&0 \\0&1 \over \sqrt{2}\\ \end{array} \right)\right] \\ Y&=\left( \begin{array}{cc} 1&-1 \\ 0&1\\ \end{array} \right) =\left[ \left( \begin{array}{cc} 1&0 \\1&1\\ \end{array} \right) , \left( \begin{array}{cc} 1 \over\sqrt{2}&0 \\0&\sqrt{2}\\ \end{array} \right)\right] . \end{align*} And since $(XY)^3 = -I$ we are done.

This answer is based on Tom Goodwill's answer on MathOverflow.

Bysshed
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Dummit and Foote gave an example of this: $(Q_8\times (\Bbb Z_2\times\Bbb Z_2))\rtimes\Bbb Z_3$. Let $i,j$ be generators of $Q_8$. $a,b$ be generators of the elementary abelian group of order $4$, and $y$ be the generator of its Sylow $3$-subgroup. $y$ acts on the Sylow $2$-subgroup by mapping $i$ to $j$, $j$ to $ij$, $a$ to $b$, and $b$ to $ab$. Now consider $i^2a$.