Showing something is a subgroup

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SUMMARY

The discussion focuses on demonstrating that a specific subset of the symmetric group, Sym(S), is a subgroup. The key element, a, must remain fixed under the bijections f and g from S to S. To establish that this subset is a subgroup, it is sufficient to show that if f(a) = a and g(a) = a, then f(g(a)) = a, confirming closure under the group operation. Additionally, it is necessary to prove that the inverse of any element in this subset also remains within the subset, specifically that f-1(a) = a.

PREREQUISITES
  • Understanding of group theory concepts, specifically symmetric groups.
  • Familiarity with bijections and their properties in set theory.
  • Knowledge of subgroup criteria, including closure and inverses.
  • Basic understanding of function notation and operations.
NEXT STEPS
  • Study the properties of symmetric groups in detail, focusing on Sym(S).
  • Learn about the criteria for subgroup determination in group theory.
  • Explore the concept of bijections and their role in group operations.
  • Investigate examples of subgroups within symmetric groups for practical understanding.
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators looking for examples of subgroup verification within symmetric groups.

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Homework Statement


Let S be a set and let a be a fixed element of S. Show that s is an element of Sym(S) such that s(a)=a is a subgroup of Sym(S).



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The Attempt at a Solution

 
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I guess I need to show associativity, having and identity, and there being an inverse, but am unsure how to.
 
The Symmetric group is already known to be a group so you do not need to show associativity. In general to show a subset H is a subgroup you need:

1) The inverse is in H
2) H is closed under the group operation (a,b in h implies ab in H).
3) If a is in H then a-1 is in H

however 2 and 3 imply 1 so you only really need to show the last two. The one step solution is to show that a in H, b in H implies ab-1 is in H because that implies 2 and 3. In your case let f,g be bijections from S to S. Then show:

1) f(a) = a implies f-1(a) =a
2) f(a) = a, g(a) =a implies f(g(a)) = a
 

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