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