Prove that a group of order 30 has at least three different normal subgroups.
There are $2$-Sylow, $3$-Sylow and $5$-Sylow subgroups. If $t_p$= number of $p$-Sylow-subgroups. Then $t_2$=$1$, $3$, $5$, $15$ and $t_3$=$1$, $10$ and $t_5$=$1$, 6. Therefore I can claim that the group is not simple. So $t_3$=1 or $t_5$=1. Which I can prove. So now we know that the group has another non-trivial normal subgroup. But this is just one normal subgroup. How can I show that there are (at least) three different normal subgroups ?