How many distinct normal subgroups H there are in the free group of rank 2 - $F_2$, so that $F_2/H \cong V_4$, where $V_4$ stands for the Klein four-group?
Normal subgroups of F2
2
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abstract-algebra
group-theory
free-groups
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0Is this the free _abelian_ group, or the free _general_ group? (It might matter, it might not.) Also, have you been able to find one? Or is this a complete mystery? Have you, for instance, tried a subgroup and gotten the cyclic group of four elements instead, or something? – 2017-02-03
2 Answers
3
Let $x$ and $y$ be the generators and let $H$ be any such group. In the quotient, all elements are $2$-torsion, so $x^2 \in H$, $y^2 \in H$, $(xy)^2 \in H$. The quotient is abelian, so $xyx^{-1}y^{-1} \in H$. But $$ \{x, y \hspace{2pc} | \hspace{1pc} x^2, y^2, (xy)^2, xyx^{-1}y^{-1}\} $$ is already the Klein $4$ group, so there is only one such subgroup $H$.
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0+1 Just a very minor remark: you may safely skip the last relation. – 2017-02-03
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0@AndreasCaranti excellent point! – 2017-02-03
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0That seems to do the trick, thank you very much! – 2017-02-03
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There is exactly one such subgroup $H$ of $F_2$, because for any two surjective morphisms $f,g\colon F_2\to C_2\times C_2=V_4$ we have $\ker(f)\cong \ker(g)$, and $F_2/\ker(f)\cong F_2/\ker(g)\cong V_4$. So then $F_2/H\cong V_4$ implies $H\cong \ker(f)$ for any such $f$.
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0This is true, but does not actually prove the theorem. It is always true that if $f, g: F_n\twoheadrightarrow G$ where $G$ is an arbitrary group generated by $\leq n$ elements then $\ker(f)\cong\ker(g)$. However, in general $\ker(f)\neq\ker(g)$. (In fact, something stronger holds: If $f: F_n\twoheadrightarrow G_1$ and $g: F_n\twoheadrightarrow G_2$ with $|G_1|=|G_2|$ then $\ker(f)\cong\ker(g)$; this holds by the Nielsen–Schreier formula if $G_1, G_2$ are finite groups.) – 2017-02-03