Every group of order $105$ has normal Sylow $5$- and $7$-subgroups and is not simple

Given: The hypotheses and objects in the Statement.

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

A Sylow $p$-subgroup of a finite group is normal if and only if it is the unique Sylow $p$-subgroup. (A Sylow $p$-subgroup is normal if and only if it is unique).

If $H\le G$ and $N⊴G$, then $HN$ is a subgroup and $H\cap N⊴H$. Here $HN:=\{hn:h\in H,n\in N\}$. (If $H\le G$ and $N⊴G$, then $HN$ is a subgroup and $H\cap N⊴H$).

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

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).

If $H\le G$ and $G$ is finite, then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$; in particular $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

A set of cardinality $n$ has exactly $n!$ bijections to itself; in particular the symmetric group on three points has order $3!=6$. (A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

For $H\le G$, the core $K={\mathrm{Core}}_G(H)$ is a normal subgroup of $G$, satisfies $K\le H$, and contains every normal subgroup of $G$ that is contained in $H$. (${\mathrm{Core}}_G(H)$ is the largest normal subgroup of $G$ contained in $H$).