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The word problem for a finitely generated free group is solvable by free reduction
Statement
Let be a free group on a finite set . The word problem in is solvable: a word on represents the identity if and only if its free reduction is the empty word.
Facts & Assumptions
Given: A finite basis and a word on .
A word is reduced exactly when no adjacent inverse pair remains, and free reduction is obtained by repeatedly deleting such pairs. (Words in an alphabet with formal inverses, elementary cancellation, and reduced words)
Every class in the reduced-word model of the free group contains exactly one reduced word. (Every class in contains exactly one reduced word)
Proof
Repeatedly apply the elementary cancellations of [L1] until no adjacent inverse pair remains. Because each cancellation shortens the word by two letters, the process halts after finitely many steps with a reduced word .
The free group element represented by is the same as that represented by , because step 1.1 used only the free-equivalence moves of [L1]. By [L2], the identity class has exactly one reduced representative, namely the empty word. Therefore represents the identity if and only if is empty.
The halting free-reduction procedure of step 1.1 therefore decides the word problem in .
Depends on
Used by
Dependency tree · two levels
9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- John Meier, Groups, Graphs and Trees (standard reference, not scraped)
- Dexter Chua after H. Wilton, Topics in Geometric Group Theory (standard reference, not scraped)