Every group of order $30$ has normal Sylow $3$- and $5$-subgroups and is not simple

Given: The hypotheses and objects in the Statement.

Let $G$ be a finite group and let $p$ be prime. If $p\mid \mathrm{\mid }G\mathrm{\mid }$, then $G$ contains an element of order $p$. (Cauchy's theorem: if a prime $p$ divides $\mathrm{\mid }G\mathrm{\mid }$, then $G$ has an element of order $p$).

For $H\le G$, left multiplication gives a transitive action on $G\mathrm{/}H$ and a homomorphism $\rho :G\to \mathrm{Sym}(G\mathrm{/}H)$ with $\ker \rho ={\mathrm{Core}}_G(H)\mathrm{.}$ (Left multiplication on $G\mathrm{/}H$ is transitive, has stabiliser $H$ at $H$, and has kernel ${\mathrm{Core}}_G(H)$).

For every natural $n$, the function $\mathrm{sgn}:S_n\to \{+1,-1\}$ is a group homomorphism. It is surjective exactly when $n\ge 2$; for $n=0$ and $n=1$ its image is $\{1\}$. (The sign is a homomorphism $S_n\to \{+1,-1\}$, surjective exactly when $n\ge 2$).

A cycle of length $k$ has sign $(-1{)}^{k-1}$. If $\sigma \in S_n$ and $c(\sigma )$ is the number of cycles after every fixed point is included as a one-cycle, then $\mathrm{sgn}(\sigma )=(-1{)}^{n-c(\sigma )}\mathrm{.}$. (A $k$-cycle has sign $(-1{)}^{k-1}$, and $\mathrm{sgn}(\sigma )=(-1{)}^{n-c(\sigma )}$ when fixed points are counted as cycles).

The image of a group homomorphism is a subgroup and its kernel is a normal subgroup. For every group homomorphism $f:G\to H$, one has $\mathrm{im}f\le H$ and $\ker f⊴G$. (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$. For every homomorphism $f:G\to H$, the rule $g\ker f\mapsto f(g)$ is an isomorphism from $G\mathrm{/}\ker f$ onto $\mathrm{im}f$. (First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$).

Let $N⊴G$. If $[G:N]$ is finite, then the quotient group $G\mathrm{/}N$ is finite and $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]\mathrm{.}$ In particular, if $G$ is finite, then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=\frac{\mathrm{\mid }G\mathrm{\mid }}{\mathrm{\mid }N\mathrm{\mid }}\mathrm{.}$. (If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$).

Let $p<q$ be primes. - If $p\nmid (q-1)$, every group of order $pq$ is cyclic. - If $p\mid (q-1)$, there are exactly two isomorphism classes of groups of order $pq$: the cyclic group $C_{pq}$ and one nonabelian semidirect product $C_q⋊C_p$. (Classification of groups of order $pq$ for primes $p<q$).

If $N⊴G$ and $P$ is a normal Sylow $p$-subgroup of $N$, then $P⊴G$. (A normal Sylow subgroup of a normal subgroup is normal in the whole group).

A group $G$ is simple if $G\mathrm{\ne }\{1\}$ and its only normal subgroups are $\{1\}$ and $G$, where normality is as in def-normal-subgroup. (Simple groups).