# Prove that a group of order 30 has at least three different normal subgroups

Prove that a group of order 30 has at least three different normal subgroups.

Prove:

$30=2\cdot3\cdot5$

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 ?

• If I understand it well, what you have proved is that the Sylow 3-subgroup and Sylow 5-subgroup are both normal, right? Call them $M$ and $N$- what can you say about $MN$? Oct 27, 2013 at 20:34
• "Therefore I can claim that the group is not simple" - why? Oct 27, 2013 at 20:35
• @Daniel Fischer: I understand, but I want to know how OP proves this. :-) Oct 27, 2013 at 20:39
• @BorisNovikov Point. Oct 27, 2013 at 20:40
• Well, claim that $t_3$=1 or $t_5$=1. Contradiction: If $t_5$=6; there are 6$\cdot$(5-1)=24 elements of order 5. And therefore $t_3$=10; there are 10$\cdot$(3-1)=20 element. 20+24>30. Contradiction, so $t_3$=1 or $t_5$=1. There is a non-trivial normal group. Thus not a simple group.
– iJup
Oct 27, 2013 at 20:49

Answer with hints that must be checked and solved:

As already proved, a group $\;G\;$ of order $\;30\;$ must have either one unique subgroup of order five or one unique subgroup of order three.

Take now the unique such subgroup, say $\;N\;$ , and one of the (possibly several) groups of the other order, say $\;H\;$ . Since clearly $\;N\cap H=1\;$ , we get that $\;NH\;$ is a subgroup (why?) of order $\;15\;$ and thus $\;NH\lhd G\;$ (why?).

But a group of order $\;15\;$ is necessarily cyclic (why?), and since any subgroup of a normal cyclic subgroup of a big group is normal in the big group (why?), we get that both $\;N,H\lhd G\;$

Let $|G|=30$. Assume that it has no nontrivial normal subgroups. Than there must exist more than one 5-sylow subgroup and more than one 3-sylow subgroup. By the sylow theorems, you can then prove that there must be 10 3-sylow subgroups and 6 5-sylow subgroups. These subgroups intersect trivially, i.e., their intersection equals {$e$}.This is true because for every subgroup every element generates the group, so if one element is in two subgroups, then the subgroups are equal. The total number of elements in $G$ however is at least the number of distinct elements in all subgroups; but $10\cdot(3-1)+6\cdot(5-1)+1>30$, which is a contradiction. QED.

• Ok, thanx for your answer! So the prove shows that $t_3$=1 or $t_5$=1.Am I right ? But not that both are unique Sylow-subgroups. How can I show that that is true ?
– iJup
Oct 27, 2013 at 21:12
• If there exists one p-sylow subgroup, it is only conjugated to itself and it must therefore be normal. With this proof you prove that $G$ has at least three normal subgroups, so indeed $t_3=1$ or $t_5=1$. You do not need to prove that they are both normal. Could you specify your question any further? Oct 27, 2013 at 21:53