There are exactly two isomorphism classes of groups of order $105$

Given: The hypotheses and objects in the Statement.

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

Let $G$ be finite, let $p$ be prime, and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then $G$ has a subgroup of order $p^a$, hence a Sylow $p$-subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$-subgroup).

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 $G$ be a finite group such that the positive integer $\mathrm{\mid }G\mathrm{\mid }$ is prime. Then every $g\mathrm{\ne }e$ has order $\mathrm{\mid }G\mathrm{\mid }$, satisfies $⟨g⟩=G$, and hence generates $G$. In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism. For each $g\in G$, the map $c_g:G\to G$, $c_g(x)=gx{g}^{-1}$, is an automorphism. (Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism).

The automorphisms of a group form a group under composition. (The automorphisms of a group form a group under composition).

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).

For every $n\ge 1$, $\mathrm{Aut}(C_n)\cong (\mathbb{Z}\mathrm{/}n{)}^{\times }\mathrm{.}$ If $C_n=⟨g⟩$, the unit class $[a]$ corresponds to the automorphism $g\mapsto g^a$. ( $\mathrm{Aut}(C_n)\cong (\mathbb{Z}\mathrm{/}n\mathbb{Z}{)}^{\times }$).

Euler's totient satisfies $\phi (1)=1$. If $p$ is prime (def-prime), then $\phi (p)=p-1.$. ($\phi (1)=1$, and $\phi (p)=p-1$ for every prime $p$).

Let $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ 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 $G$).

Let $N,H\le G$. The conditions $N⊴G,G=NH,N\cap H=\{1\}$ hold if and only if both of the following hold: conjugation ${\alpha }_h(n)=hn{h}^{-1}$ restricts to an action $\alpha :H\to \mathrm{Aut}(N)$, and the resulting map $\mathrm{\Phi }:N{⋊}_{\alpha }H⟶G,(n,h)⟼nh$ is an isomorphism carrying the canonical factors onto $N$ and $H$. ( Recognition theorem: $G=NH$ with $N⊴G$, $N\cap H=1$ exactly realises an external semidirect product).

The canonical factors of $N{⋊}_{\alpha }H$ form an internal direct product if and only if ${\alpha }_h={\mathrm{id}}_N$ for every $h\in H$. In that case $N{⋊}_{\alpha }H$ is the external direct product $N\times H$. (The canonical semidirect decomposition is an internal direct product if and only if the defining action is trivial).

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 $g\in G$ and $h\in H$ have finite orders $m,n\ge 1$, then in the external direct product $\mathrm{ord}(g,h)=\mathrm{lcm}(m,n)\mathrm{.}$ (If $g$ and $h$ have finite orders $m$ and $n$, then $\iota (\mathrm{ord}(g,h))=\mathrm{lcm}(\iota (m),\iota (n))$ in $G\times H$).

If $G=⟨g⟩$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong (\mathbb{Z},+)$; - if $g$ has finite order $n$, necessarily $n\ge 1$, then $G\cong (\mathbb{Z}\mathrm{/}n,+)$. (Every cyclic group is isomorphic to $(\mathbb{Z},+)$ or to $(\mathbb{Z}\mathrm{/}n,+)$ for its finite order $n\ge 1$).