Help with a hard group theory question?

Question

Here is the question I would like to answer:
Does A5 (the alternating group of degree five) contain a subgroup of order m for each factor m of 60?

My gut says yes but I have no idea how to prove this. Is there a way to do it without having to go through and find individual subgroups for each factor (since that would take a very long time)?

Thanks very much.

Answer 1

A_5 is simple, and therefore cannot contain a subgroup of order 30, which would necessarily be normal. In fact, A_5 can't contain a subgroup of order 20 or 15 either, though the reasoning is a little indirect: turns a out a simple finite group G with subgroup H of index m embeds in A_m (to prove this, consider the action of G on the cosets of H).



A_5 does have subgroups of order 12, 10, and 6. 12 and 6 are easy (A_4 and S_3), and 10 isn't too hard either. A little more exploring will show you that every group of order in A_5 is isomorphic to A_4. In fact, every group of order 12 in A_5 is a normalizer for a 2-group, every one of order 10 a normalizer for a 5-group, and every one of order 6 a normalizer for a 3-group.



Finally, there are of course subgroups of order 5, 3, and 2, considering Sylow's theorems.



As for Curt's answer, there is a flaw: products of subgroups aren't necessarily groups themselves.



Steve

Answer 2

Some thoughts of how to attack it:



1. See what the Sylow Theorems say.

2. Find cyclic subgroups of orders 2, 3, 4, and 5. Take their products and see what the orders of the resulting groups are. :)

3. Answer the same question for A4, embedded in A5 in the obvious way. For those groups, take their products with the groups generated by by (45) and (12345).

Answer 3

The identification element is often in the kernel of any homomorphism. If f is a homomorphism from G to H (the place e is the identification of G and e' is the identification of H). Then f(e) = f(ee) = f(e)f(e) Canceling f(e) from the two sides leaves you with e' = f(e). this suggests the identification of G continuously maps to the identification of H, so e could desire to be in Ker(f). showing inverses is almost basically as ordinary. think g is in Ker(f). Then f(g) = e'. So, e' = f(e) = f(gg^-a million) = f(g)f(g^-a million) = e'f(g^-a million) = f(g^-a million) this suggests that g^-a million maps to e' so it quite is in Ker(f).