Talk:Non-prime solvable fundamental groups (Ex)
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− | Suggestion: The first manifold is $M = S^3$. Other than that, non-prime manifolds possess prime decompositions. So there must be at least two manifolds connected sum to each other. The fundamental group of $M$ is the free product of the fundamental groups of the various fundamental groups. Since it's solvable, note that each abelian quotient group can only contain elements from any individual fundamental group, otherwise it wouldn't be abelian. Suppose at least two fundamental groups of the component manifolds $M_i$ are nontrivial. Consider the first time in the series $G_n \triangleright G_{n-1}$ such that $G_n$ contains a nontrivial element $x$ from some $\pi_1(M_i)$, where no element from $\pi_1(M_i)$ appears in $G_{n-1}$. $xG_{n-1}x^{-1}$ is a contradiction to normality, since they are in a free product. | + | Suggestion: The first manifold is $M = S^3$. Other than that, non-prime manifolds possess prime decompositions. So there must be at least two manifolds connected sum to each other. The fundamental group of $M$ is the free product of the fundamental groups of the various fundamental groups. Since it's solvable, note that each abelian quotient group can only contain elements from any individual fundamental group, otherwise it wouldn't be abelian. Suppose at least two fundamental groups of the component manifolds $M_i$ are nontrivial. Consider the first time in the series $G_n \triangleright G_{n-1}$ such that $G_n$ contains a nontrivial element $x$ from some $\pi_1(M_i)$, where no element from $\pi_1(M_i)$ appears in $G_{n-1}$. $xG_{n-1}x^{-1}$ is a contradiction to normality, since they are in a free product. If there's only one component $M_1$ with nontrivial fundamental group, then use the Hurewicz theorem to control the homology of the remaining components. |
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Revision as of 06:37, 8 January 2019
Suggestion: The first manifold is . Other than that, non-prime manifolds possess prime decompositions. So there must be at least two manifolds connected sum to each other. The fundamental group of is the free product of the fundamental groups of the various fundamental groups. Since it's solvable, note that each abelian quotient group can only contain elements from any individual fundamental group, otherwise it wouldn't be abelian. Suppose at least two fundamental groups of the component manifolds are nontrivial. Consider the first time in the series such that contains a nontrivial element from some , where no element from appears in . is a contradiction to normality, since they are in a free product. If there's only one component with nontrivial fundamental group, then use the Hurewicz theorem to control the homology of the remaining components.