Group with exactly six Sylow 5-subgroups?

Question

Give an example of a group with exactly six Sylow 5-subgroups.

I think $A_5$ works because it has 6 subgroups of order 5:

$\langle(12345)\rangle,\langle(12354)\rangle, \langle(12435)\rangle, \langle(12453)\rangle, \langle(12534)\rangle, \langle(12543)\rangle$. Is this right? Is there a simpler group that meets these requirements?

That is correct $|A_5| = 60 = 5*12$. So the number of $5$-Sylow subgroups is either $1$ or $6$. You just wrote down 6 $5-$Sylow subgroups. — Ken Duna, Apr 27 '16 at 19:44
@skmehta "Simple" - Yet another way to see that there is not 1 $5$-Sylow subgroup. Though explicitly listing them is nice. — Ken Duna, Apr 27 '16 at 19:46
I'm not sure if there's a simpler example, but it's the smallest example. — verret, Apr 27 '16 at 19:54

score 3 · Accepted Answer · answered Apr 27 '16 at 19:49

Your idea of $A_5$ is a good one. $|A_5| = 60 = 5*12$, so using the Sylow Theorems, $n_5 = 1$ or $n_5 = 6$ (where $n_5$ is the number of $5$-Sylow subgroups). $n_5 \neq 1$ since if $n_5=1$, you would have that the $5$-Sylow subgroup is normal, a contradiction of the simplicity of $A_5$.

Therefore $n_5 = 6$.

In your question statement you explicitly listed the subgroups, which I personally like as a solution since it does not require you to use the fact that $A_5$ is simple.

Group with exactly six Sylow 5-subgroups?

1 Answers1