Proving H as a Subgroup of S5 | Abstract Algebra Help

[JasonJo]

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.

 

[StatusX]

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.

 

[JasonJo]

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?

 

[StatusX]

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.

 

[JasonJo]

edit:

so ridiculously lost

 

[StatusX]

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.

 

[JasonJo]

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?

 

[StatusX]

What element a has a^2=(12)(34)?

 

[JasonJo]

oooh thanks!