Calculating the commutator subgroup.

Asked Dec 27 '13 at 10:09

Active Dec 27 '13 at 10:09

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I want to find the commutator subgroup ($G^k$, $k=1,2,..$) of such matrix group:$$ \left(\begin{array}{ccc} 1 & * & * \\ 0 & * & 0 \\ 0 & * & * \\ \end{array}\right)$$

I have multiplied them "by hands" and I got that $G^1= \left(\begin{array}{ccc} 1 & * & * \\ 0 & 1 & 0 \\ 0 & * & 1 \\ \end{array}\right)$, $G^2= \left(\begin{array}{ccc} 1 & * & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right)$, $G^3=E$.

Did I do it in the right way? Or maybe you can suggest a more elegant way of calculating?

asked Dec 27 '13 at 10:09

xxxxx

CAn you more explain that what is exactly your matrix group and over what?Natural number or $\mathbb Z$? – Bobby Dec 27 '13 at 10:57
2

That's right for matrices over any field except ${\mathbb F}_2$, where $G^2=1$. – Derek Holt Dec 27 '13 at 11:37
Would you please give a definition of your group and notation? – Phira Dec 27 '13 at 13:43

Calculating the commutator subgroup.

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