Define $\displaystyle A_n$ as the alternating group. How do I figure out how many 5-cycles $\displaystyle A_5$ has?

Printable View

- May 6th 2009, 09:33 AMJason BourneGroups - Alternating group question
Define $\displaystyle A_n$ as the alternating group. How do I figure out how many 5-cycles $\displaystyle A_5$ has?

- May 6th 2009, 09:45 AMNonCommAlg
- May 6th 2009, 01:51 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 $\displaystyle A_5$ ?

(ii) How many Sylow 5-subgroups does $\displaystyle A_5$ have?

For (i) I think the answer is 5 because $\displaystyle 60=2^{2}.3.5^{1}$

for (ii) I think using one of the Sylow theorems, that the number of Sylow 5-subgroups is of the form $\displaystyle 5k+1$ and divides the order of $\displaystyle A_5$, so I guess from this there is either 1 or 6 Sylow 5-subgroups in $\displaystyle A_5$, so the answer is 6?? Can anyone verify any of this?

(iii) If $\displaystyle P$ is a Sylow p-subgroup of a finite group $\displaystyle G$, and if $\displaystyle Q$ is a Sylow p-subgroup of a finite group $\displaystyle H$, prove that the direct product $\displaystyle P \times Q$ is a Sylow p-subgroup of the direct product $\displaystyle G \times H$.

For (iii) i attempted to prove it like follows: Take $\displaystyle (p_1,q_1),(p_2,q_2) \in P \times Q $ $\displaystyle (p_1,p_2 \in P, q_1,q_2 \in Q)$

$\displaystyle (p_1,q_1)(p_2^{-1},q_2^{-1}) = (p_1p_2^{-1},q_1q_2^{-1}) \in P \times Q $

Is this the right method? Thanks for any help anyone can provide. - May 6th 2009, 02:04 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 $\displaystyle |P|=p^m, \ |Q|=p^n, \ |G|=p^m r, \ |H|=p^n s,$ where $\displaystyle \gcd(p,r)=\gcd(p,s)=1.$ then $\displaystyle |P \times Q|=p^{m+n}$ and

$\displaystyle |G \times H|=p^{m+n}rs$ and clearly $\displaystyle \gcd(rs,p)=1.$ this completes the proof. - May 7th 2009, 12:34 AMTheAbstractionist
Hi

**Jason Bourne**.

If you are allowed to assume that $\displaystyle A_5$ 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 $\displaystyle A_5;$ since $\displaystyle A_5$ is simple, there must therefore be more than 1 Sylow 5-subgroup (indeed more than 1 Sylow subgroup of any order). - May 7th 2009, 04:11 AMJason Bourne
- May 7th 2009, 03:01 PMTheAbstractionist
Hi

**Jason Bourne**.

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

Think about it. (Wink) - May 8th 2009, 11:35 PMJason Bourne