Proving H as a Subgroup of S5 | Abstract Algebra Help

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Homework Help Overview

The discussion revolves around proving that a set H is a subgroup of S5, the permutation group of 5 elements, and exploring the properties of a set defined as H={g^2:g is an element of G} in the context of group theory.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of H in the context of subgroup criteria and question whether the argument holds for larger groups. There is exploration of whether H is a subgroup under various conditions, particularly focusing on abelian versus non-abelian groups.

Discussion Status

Multiple interpretations of H are being explored, and some participants are attempting to find counterexamples to support their claims. Guidance has been offered regarding the properties of non-abelian groups and the implications for subgroup status.

Contextual Notes

There is uncertainty regarding the specific elements of H and the conditions under which the subgroup properties hold. Participants express confusion about finding appropriate examples and the implications of group order on subgroup characteristics.

JasonJo
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1) prove that H is a subgroup of S5 (the permutation group of 5 elements). every element x in H is of the form x(1)=1 and x(3)=3, meaning x moves 1 to 1 and moves 3 to 3. does your argument work hen 5 is replaced by a number greater than or equal to 3?

2) Let G be a group. prove or disprove that H={g^2:g is an element of H} is a subgroup.
 
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What exactly is H in 1?

For 2, note that it is a subgroup if G is abelian, but the proof of this should show there's no reason to expect it to be true in general, and it shouldn't be hard to find a counterexample.
 
StatusX said:
What exactly is H in 1?

For 2, note that it is a subgroup if G is abelian, but the proof of this should show there's no reason to expect it to be true in general, and it shouldn't be hard to find a counterexample.

i actually fouund that out also but I am having a very hard time finding an example. maybe I am just not seeing it?
 
Just look at some non-abelian groups. You only need to find some elements a and b of G such that aabb is not the square of any element of b. Note a must not commute with b. S_3 is the smallest non-abelian group, but the associated H is A_3, a subgroup, so this won't work. Try A_4, S_4, etc.
 
edit:

so ridiculously lost
 
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Well, I said the smallest non-abelian group, S_3, won't work. The next smallest is A_4, and it does work. One way to prove this is to show the size of the set of elements which are squares does not divide the order of the group. EDIT: Not that it really matters, but, I realized A_4 isn't the next smallest. There are the dihedral groups, etc.
 
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i get that |G| = 12 and |H| = 12

i have that A4={e, (123), (234), (341), (412), (132), (243), (314), (421), (12)(34), (13)(24), (14)(23)}

then i get that H= {e, (132), (243), (314), (421), (123), (234), (341), (12)(34), (13)(24), (14)(23)}

or i get that G= G^2 = H

what am i doing wrong?
 
What element a has a^2=(12)(34)?
 
oooh thanks!
 

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