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Let M be the squares space $(R\times Z) \cup (Z\times R)$ covering the 8 space (pointwise union of two circles) by calling one circle of the 8 (path starting and ending at the dot of intersection of the two circles) x and the other y, and declaring each path from one dot to the one right to it as covering x and each path from one dot to the one above it as covering y. Call the cover map p and let there be $p_*:\pi(M_1)\rightarrow\pi_1(8) : \varphi\mapsto p\circ\varphi$. Show that $J=p_*(\pi_1(M))$ Is the commutator subgroup of $\pi_1(8)=F_2$. I have shown that the commutator subgroup is contained in J but I am struggling with the other direction. I have many ideas:

  • I know that $AutM=Z^2=F2/F2'=N(J)/J=F2/J$, but I'm not sure that from this follows that $J=F2'$.
  • I have discovered that $J=\{x^{i_1}y^{j1}x^{i_2}...| \sum i_k=\sum j_k=0\}$, from which it is obvious that $F2' \subseteq J$ But I am not sure how to show the other way. Wikipedia says the commutator subgroup can be thought of as a group of $g_1...g_n$ such that by rearranging the g's we get the identity, which means I'm right, I just don't know how to prove it.
  • I have shown manually that simple pathes are commutators (using the fact that a loop around each square is a commuator) but this proof failes for complex, self-intersecting paths.

Help will be appreciated.

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1 Answer 1

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Let $s : F_2 \to F_2/F_2'$. Since $F_2' \subseteq J \subseteq F_2$, $s(J) = J/F_2'$ is a subgroup of $F_2/F_2'$, which is abelian. Now, since $J = \{x^{i_1}y^{j_1}x^{i_2}\ldots \}$ where $\sum i_k = \sum j_k = 0$, $s(J) = \{s(x)^{i_1}s(y)^{j_1}s(x)^{i_2}\ldots \}$. Since $F_2/F_2'$ is abelian, $s(J) = \{s(x)^{i_1}s(x)^{i_2} \ldots s(y)^{j_1}s(y)^{j_2}\ldots\} = \{s(x)^{i_1+i_2+\ldots} s(y)^{j_1+j_2+ \ldots} \} = \{s(x)^0 s(y)^0 \} = \{1\}$.

You can also show more straightforwardly, how $x^{i_1}y^{j_1}x^{i_2}\ldots = (x^{i_1}y^{j_1}x^{-i_1}y^{-j_1})(y^{j_1}x^{i_1+i_2}y^{-j_1}x^{-i_1-i_2})(x^{i_1+i_2}y^{j_1+j_2}x^{-i_1+i_2}y^{-j_1+j_2}) \ldots$ with some leftover at the end $x^{\sum i_k}y^{\sum j_k}$, or $y^{\sum j_k}x^{\sum i_k}$ depending on the parity of the length of the sequence. But if you pick a sequence in $J$, those leftovers are trivial, so there you have the explicit way to write your element in $J$ as a composition of commutators from $F_2$

And thus $J = F_2'$

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The first solution uses algebra not at all straightforward for me, though I understood it eventually! Also, the second solution is just what I was looking for, apparently I just didn't try hard enough... Thank you very much! –  Idan Jan 26 '13 at 18:03

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