This is about Sylow subgroups of Alternating group $A_n$ (Multiple choice)

Question

This is a question from a competitive exam.

For a positive integer $n\ge 4$ and a prime number $p\le n$ denote $U_{p,n}$ to be the union of all $p$-sylow subgroups of alterbating group $A_n$. Also let $K_{p,n}$ denote the subgroup of $A_n$ generated by $U_{p,n}$ and let $|K_{p,n}|$ denote the order of $K_{p,n}$.Then

1. $|K_{2,4}|=12$
2. $|K_{2,4}|=4$
3. $|K_{2,5}|=60$
4. $|K_{3,5}|=30$

It may have more than one correct answer.

I don't know how to start, I am completely stuck.Please help.

Thnx in advance.

Answer 1

You just have to check the answers case-by-case. My writing here was a bit of a mess, so here are full solutions. Just read the first solution and don't look at (3) and (4) if you want to solve them yourself (they use the same ideas)

1. $A_4$ has order $12$, so a $2$-Sylow has order $4$. The $2$-Sylow is the Klein $4$-group $\{(),(12)(34),(13)(24),(14)(23)\}$. It's normal (you can check this directly), so must equal $K_{2,4}$, since all $2$-Sylows are conjugate.

2. See (1)

3. Take the $2$-Sylow of $A_4$ from (1). It fixes the element $5$. Furthermore, there are $2$-Sylows fixing $1,2,3,4$, so there are $5$ $2$-Sylows that only share the unit element. Thus $K_{2,5}$ has at least $1+5*3=16$ elements and must divide $60$ and be divisible by $4$. Since $K_{2,5}$ contains the product $((12)(34))((15)(34)) = (152)$ it contains elements of order $3$, so can't have order $20$. Therefore, $K_{2,5}=A_5$.

4. $A_5$ is simple, so it has no non-trivial normal subgroups. A subgroup of index $2$ must be normal, so this is impossible. This is trivial to show: Let $(G:H)=2$, since cosets cover the whole set, i.e. $H\cap xH=H\cap Hx=\emptyset$ and $G = H\cup xH = H\cup Hx$ for $x\not\in H$ we have $xH=Hx$, so $H$ is normal in $G$. This can furthermore be generalized, so that if $(G:H)=p$, where $p$ is the smallest prime dividing $|G|$, then $H$ is normal (look at the conjugation action of $G$ on $H$, which gives a homomorphism into $S_p$ and look at the kernel...)

Answer 2

The more standard notation for $K_{p,n}$ is $O^{p'}(A_n)$. It is the smallest normal subgroup whose index is not divisible by $p$. For each one you need to decide if (1) the number is the order of a normal subgroup, (2) the index is not divisible by $p$, and (3) if it is the smallest number like that. For $n=4$, this is a very small group, so just list all elements of $U_{p,n}$. For $n \geq 5$, $A_n$ has very few normal subgroups so the answer is fairly simple.

The complete answer is:

$$K_{p,n} = \begin{cases} 1 & \text{ if } \max(p,3) > n \text{ or } (p,n) = (2,3)\\ A_n & \text{ if } \max(p,5) \leq n \text{ or } (p,n) = (3,3) \text{ or } (p,n) = (3,4) \\ K_4 & \text{ if } (p,n) = (2,4) \end{cases}$$

where $K_4$ is the Klein four-group.

This says: If $n$ is too small, then $K_{p,n}=1$ since $A_n$ has only the identity Sylow $p$-subgroup. If $n$ is big enough, $K_{p,n}=A_n$, since it is a non-identity normal subgroup. On the border you have to check cases, and $(p,n)=(2,4)$ is actually different.