Group of order 30 can't be simple I have this following question from my class note on Sylow Theorem:

Show that a group of order 30 can not be simple.

For that I know the followings:
(1) A simple group is one that does not have non-trivial normal group, 
(2) Group $G$ is p-group if there exists an integer $e$ such that $|G| = p^e,$ 
(3) A $p$-subgroup $H$ of $G$ is called Sylow p-subgroup of $G$ if $p \nmid |G/H|,$ and finally 
(4) A subgroup $H$ of $G$ is a Sylow p-subgroup if there exist integers $e$ and $m$ such that $|G| = p^em,$ and $(e, m) = 1,$ and also $|H| = p^e,$
but unfortunately I don't know how to weave them into a solution, any help would therefore be very much appreciated. Thank you for your time and help.
PS. If you have choices of elegant versus dummy down-to-earth solutions, do please give me the latter. I know it would be tedious for you but you have a slowpoke turtle down here. Thank again for your patience.
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I came across a very simple solution by John Beachy here, consisting of only 2 lines, which I adapted to fit group order of 30:
EDIT (j.p.) Unfortunately the linked solution cannot be easily adapted to this case here. I kept the text as far it was correct.
(1) Recall Sylow's Theorem: If $|G| = p^em; p, e, m \in \mathbb Z^+; (p, m) = 1;$ the number of $Syl_p (G) = s$, then $s \equiv 1\pmod p$ and $s \mid m.$ 
(2) For $|G| = 5 \cdot 2 \cdot 3$, suppose that $p = 5, e = 1$ and $m = 6.$ Therefore $s \equiv 1\pmod 5 $ and $s \mid 6,$ thus $s = \{1, 6 \}$. The 5-Sylow subgroup is normal if and only if $s=1$, and in this case $G$ is not simple.
(3) Now look at Nirdonkey's answer.
Hope this will be useful for some of you and thank you very much for those of you who have helped me. 
 A: By Cayley's theorem, a group $G$ of order $30$ is isomorphic to a subgroup of $S_{30}$ is such a way that no non-identity element of $G$ has any fixed point. There is an element $t$ of order $2$ in $G$ which is represented by a product of $15$ $2$-cycles, so as an odd permutation. The elements of $G$ which are represented as even permutations of $S_{30}$ form a normal subgroup of index $2$, so $G$ is not simple.
A: If $G$ is a group of order 30, then the number of Sylow 5-subgroups is congruent to 1 modulo
5 and divides 30, so it must be 1 or 6. And the number of Sylow 3-subgroups is congruent to
1 modulo 3 and divides 30, so it must be 1 or 10. If there is a unique subgroup of order 3 or
5, it is normal and we are done. If not, there are 6 subgroups of order 5 and 10 subgroups of
order 3. But all of these groups must be cyclic, so every non-trivial element of such a subgroup
is a generator of the subgroup. It follows that no non-trivial element of $G$ is contained in more
than one of these subgroups. But then we have at least $6 \cdot 4 + 10 \cdot 2$ elements in $G$, which is a
contradiction.
See also Group of order $|G|=pqr$, $p,q,r$ primes has a normal subgroup of order
