Define as the alternating group. How do I figure out how many 5-cycles has?

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- May 6th 2009, 10:33 AMJason BourneGroups - Alternating group question
Define as the alternating group. How do I figure out how many 5-cycles has?

- May 6th 2009, 10:45 AMNonCommAlg
- May 6th 2009, 02:51 PMJason Bourne
Thank you for your post. There are a few more questions concerning this and Sylow theorem that maybe you can help me with.

(i) What is the order of every Sylow 5-subgroup of ?

(ii) How many Sylow 5-subgroups does have?

For (i) I think the answer is 5 because

for (ii) I think using one of the Sylow theorems, that the number of Sylow 5-subgroups is of the form and divides the order of , so I guess from this there is either 1 or 6 Sylow 5-subgroups in , so the answer is 6?? Can anyone verify any of this?

(iii) If is a Sylow p-subgroup of a finite group , and if is a Sylow p-subgroup of a finite group , prove that the direct product is a Sylow p-subgroup of the direct product .

For (iii) i attempted to prove it like follows: Take

Is this the right method? Thanks for any help anyone can provide. - May 6th 2009, 03:04 PMNonCommAlg
your answers to (i) and (ii) are correct, although you need to explain in (ii) why the number of Sylow 5-subgroups cannot be 1. for that you need to look at your first question about the number

of 5 cycles. for (iii) look at the orders: since P and Q are Sylow p-subgroups we have where then and

and clearly this completes the proof. - May 7th 2009, 01:34 AMTheAbstractionist
Hi

**Jason Bourne**.

If you are allowed to assume that is simple, then the answer is immediately obvious. If there were just 1 Sylow 5-subgroup, this would be a proper nontrivial normal subgroup of since is simple, there must therefore be more than 1 Sylow 5-subgroup (indeed more than 1 Sylow subgroup of any order). - May 7th 2009, 05:11 AMJason Bourne
- May 7th 2009, 04:01 PMTheAbstractionist
Hi

**Jason Bourne**.

Are you aware of the result that for each prime dividing the order of a finite group all Sylow -subgroups of are conjugate to each other? It follows that if has a unique Sylow -subgroup, then that Sylow -subgroup is normal in

Think about it. (Wink) - May 9th 2009, 12:35 AMJason Bourne