Is the Generated Set of a Group Closed Under Group Operation?

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SUMMARY

The subgroup H generated by the elements α = (1 2) and β = (2 3) in the symmetric group S3 is not closed under the group operation. The generated set H = {ε, (1 2), (2 3), (1 2 3), (1 3 2)} fails to include the element (1 3), which is produced by the operation (1 2 3)(1 2). This indicates that while H contains several elements, it does not encompass all possible products of α and β, thus violating the closure property essential for a subgroup.

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  • Understanding of group theory concepts, specifically subgroups and closure properties.
  • Familiarity with symmetric groups, particularly S3.
  • Knowledge of cycle notation for permutations.
  • Basic operations within group theory, including multiplication of group elements.
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  • Study the properties of symmetric groups, focusing on S3 and its subgroups.
  • Learn about the closure property in group theory and its implications for subgroup formation.
  • Explore the concept of generating sets and how to determine if a set is closed under group operations.
  • Investigate the relationship between cycle types and subgroup generation in permutation groups.
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Students of abstract algebra, mathematicians specializing in group theory, and anyone interested in the properties of symmetric groups and their substructures.

BrockDoiron
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Hi, I am told to give the subgroup H=<α,β> with α,β[itex]\in[/itex]S3

α = (1 2)
β = (2 3)

So I know that H={αkβj|j,k[itex]\in[/itex](the integers)}
However, would αβα or βαβ (in this case, they're equal) be in H?

The set H={ε,(1 2), (2 3), (1 2 3), (1 3 2)} (or {ε,α,β,αβ,βα})
would not be closed because (1 2 3)(1 2) = (1 3) which is not in H
But if (1 3) is in H you have all of S3 which I thought was only generated by a 2-cycle and a 3-cycle.
 
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Yes, a group is closed under the group operation so any combinations of [itex]\alpha[/itex] and [itex]\beta[/itex] must also be in the group.
 

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