question on Sylow groups

For a positive integer $n\ge 4$ and a prime $p\le n$, let $U_{p,n}$ denote the union of all $p$-Sylow subgroups of the alternating group $A_n$ on $n$ letters. Also let $K_{p,n}$ denote the subgroup of
$A_n$ generated by $U_{p,n}$. Then the order of $K_{2,4}=5$ and order of $K_{2,5}=60$.

It is given in a question paper of an examination for $\mathbf{NET}$. I can’t do it. If someone knows, please help me.

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For $A_4$: You can just list down all 12 elements and you will see that there is a unique 2-Sylow subgroup $H = \{e, (12)(34), (13)(24), (14)(23)\}$. Hence, $U_{2,4} = K_{2,4} = H$ so $|K_{2,4}| = 4$.

For $A_5$ : You are looking at subgroups of order 4. Since any group of order 2 is contained a 2-sylow subgroup, you can just count the elements of order 2. $A_5$ has 15 elements of order 2 of the form $(12)(34), (12)(35)$, etc. Hence, $|U_{2,5}| \geq 15+1 = 16$ and so $|K_{2,5}| \in \{20, 30, 60\}$ (since it must divide 60)

Assume $|K_{2,5}| = 20$, then let $A_5$ act on $A_5/K_{2,5}$ by left translation, and this will give you a homomorphism $A_5 \mapsto S_3$. Since $A_5$ is simple, this homomorphism must be injective, which means
60 = |A_5| \leq |S_3| = 6
Hence, $|K_{2,5}| \neq 20$.

Again, if $|K_{2,5}| = 30$, then it has index 2, and so has to be normal in $A_5$, which is impossible. Thus, $|K_{2,5}| = 60$.

If you know that $A_5$ is simple and that all Sylow subgroups are conjugated, then this question is simple: since the set of all Sylow subgroups is invariant under conjugation, they generate a normal subgroup. So $K_{2,5}=A_5$ has order $60$. In fact, this will work more generally for finite simple groups and any prime $p$ dividing their order, for instance $A_{19}$ is generated by its Sylow 11-subgroups.

So let’s say you don’t know that $A_5$ is a simple group. Then instead you can observe that $K_{2,5}$ contains all permutations of the type $(a\ b)(c\ d)$, with $a, b, c, d$ all different, since these elements have order $2$ and elements of order $2^i$ are always contained in (at least one) Sylow $2$-subgroup. From the equality
$$(a\ b)(b\ c) = (a\ b)(d\ e)\cdot (d\ e)(b\ c),$$
if follows that we can use these elements to generate all elements of type $(a\ b)(b\ c)$, since we can always find two extra elements $d,e$ different from $a,b,c$. (This step fails in $A_4$!)

But since every element of $A_5$ is generated by an even number of involutions, we can write an arbitrary element $x$ of $A_5$ as a product of elements of type $(a\ b)(c\ d)$ or $(a\ b)(b\ c)$. Since each of these is contained in $K_{2,5}$, so is $x$ and we are done.

(This argument holds in every alternating group $A_n$, $n\geq 5$. But it only works for Sylow $2$-subgroups.)