Find all the Sylow $2$ and the Sylow $3$ subgroups of $A_5$.

Question

Find all the Sylow $2$ and the Sylow $3$ subgroups of $A_5$.

My try:

$|A_5|=60$.Number of Sylow $2$ subgroups of $A_5=n_2=1,3,5,15$.

Number of Sylow $3$ subgroups of $A_5=n_3=1,4,10$.Neither $n_2$ nor $n_3=1$ as $A_5$ is simple.

Now the Sylow $2$ subgroups of order $4$ may be:

• $\{e,(12)(34),(14)(23),(13)(24)\}$,
• $\{e,(15)(34),(13)(45),(14)(35)\}$
• $\{e,(23)(45),(24)(35),(25)(34)\}$
• $\{e,(12)(35),(13)(25),(15)(23)\}$
• $\{e,(12)(45),(15)(24),(14)(25)\}$

I am wondering if there are any other Sylow 2 subgroup of order $4$.

Regarding the Sylow $3$ subgroups ;

Number of $3$ cycles in $A_5=20$ .Each of these cycles gives rise to a subgroup of order $3$ given by

• $\{e,(123),(132)\}$
• $\{e,(143),(134)\}$ and so on .As each subgroup has two elements of order $3$,so we get $10$ subgroups.

I think it is correct in case of $A_5$.But I am not sure in case of $A_4$ .Please help.

Answer 1

Note that as $A_5$ is simple, there are more than one Sylow 5-subgroups. Namely there are $6$. As we know that there are $10$ Sylow 3-subgroups. Then at least $44$ permutations have a order $5$ or $3$. Hence we're left with $15$ more element. As every Sylow 2-group has at least $2$ unique elements it's impossible to have $15$ Sylow 2-subgroups. As you've listed all $5$, then number of Sylow 2-subgroups is exactly $5$.