Group of order 30 can't be simple

Solution 1:

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

Solution 2:

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.