The p-prime core of a finite group

Let $G$ be a finite group (Group and abelian group, The cardinality $\mid A\mid$ of a finite set) and let $p$ be a prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$).

$p$-element terminology. An element $g\in G$ of finite order is a $p$-element when $\mathrm{ord}(g)$ is a power of $p$, and a $p^{\mathrm{\prime }}$-element when $p\nmid \mathrm{ord}(g)$ (The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity, A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$). A subgroup $K\le G$ is a $p$-subgroup when $\mathrm{\mid }K\mathrm{\mid }$ is a power of $p$, and a $p^{\mathrm{\prime }}$-subgroup when $p\nmid \mathrm{\mid }K\mathrm{\mid }$; the trivial subgroup is both. The terminology is used without further comment throughout the normal-complement material of this page.

The $p^{\mathrm{\prime }}$-core. Call a subgroup $N\le G$ of $G$ a normal $p^{\mathrm{\prime }}$-subgroup when $N⊴G$ and $p\nmid \mathrm{\mid }N\mathrm{\mid }$ (Normal subgroup: invariance under conjugation). The $p^{\mathrm{\prime }}$-core of $G$ is

$O_{p^{\mathrm{\prime }}}(G):=⟨\bigcup \{N\le G:N⊴G,p\nmid \mathrm{\mid }N\mathrm{\mid }\}⟩,$

the subgroup generated by all normal $p^{\mathrm{\prime }}$-subgroups of $G$ (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups). It is the unique largest normal $p^{\mathrm{\prime }}$-subgroup of $G$: it is normal in $G$, its order is prime to $p$, and every normal $p^{\mathrm{\prime }}$-subgroup of $G$ is contained in it.

Why the definition is well posed. The family $\mathcal{S}=\{N\le G:N⊴G,p\nmid \mathrm{\mid }N\mathrm{\mid }\}$ is nonempty ($\{1\}\in \mathcal{S}$, since $p\nmid 1$) and finite: every member is a subset of the finite set $G$, and $G$ has finitely many subsets (The cardinality $\mid A\mid$ of a finite set, $\mid \mathcal{P}(A)\mid =2^{\mid A\mid }$ for finite $A$). Write $\mathcal{S}=\{N_1,\dots ,N_r\}$.

Products stay in the family. If $M,N\in \mathcal{S}$, then $MN$ is a subgroup of $G$ (If $H\le G$ and $N⊴G$, then $HN$ is a subgroup and $H\cap N⊴H$), it is normal because $g(MN)g^{-1}=gM{g}^{-1}gN{g}^{-1}=MN$ for every $g\in G$ (Normal subgroup: invariance under conjugation, Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism), and its order divides $\mathrm{\mid }M\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }$: by the second isomorphism theorem $M\mathrm{/}(M\cap N)\cong MN\mathrm{/}N$, so $\mathrm{\mid }MN\mathrm{\mid }=\mathrm{\mid }M\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }\mathrm{/}\mathrm{\mid }M\cap N\mathrm{\mid }$ (Second isomorphism theorem for groups: $H\mathrm{/}(H\cap N)\cong HN\mathrm{/}N$, 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 divisor of the $p^{\mathrm{\prime }}$-number $\mathrm{\mid }M\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }$ is again prime to $p$: if the prime $p$ divided $\mathrm{\mid }MN\mathrm{\mid }$ it would divide $\mathrm{\mid }M\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }$ and hence, by Euclid's lemma, one of $\mathrm{\mid }M\mathrm{\mid }$, $\mathrm{\mid }N\mathrm{\mid }$ (Euclid's lemma: if $p$ is prime and $p\mid ab$ then $p\mid a$ or $p\mid b$, Divisibility is reflexive and transitive on $\mathbb{Z}$, and is linear: if $d\mid a$ and $d\mid b$ then $d\mid ax+by$ for all integers $x,y$; also $d\mid a$ implies $d\mid ac$, $-d\mid a$ and $d\mid -a$). So $MN\in \mathcal{S}$.

The generated subgroup is a member. By the product closure just proved, $T:=N_1\cdots N_r$ belongs to $\mathcal{S}$, by finite induction. This subgroup contains each $N_i$ (insert identities in all other factors), hence contains $⟨\bigcup \mathcal{S}⟩$ by the defining minimality of the generated subgroup (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups). Conversely every factor $N_i$ lies in that generated subgroup, so their product $T$ does too. Thus $O_{p^{\mathrm{\prime }}}(G)=T\in \mathcal{S}$; in particular $O_{p^{\mathrm{\prime }}}(G)=\bigcup \mathcal{S}$, because $T$ is itself a member of the family and contains every member.

Largest and unique. As a member of $\mathcal{S}$, $O_{p^{\mathrm{\prime }}}(G)$ is a normal $p^{\mathrm{\prime }}$-subgroup of $G$, and it contains every $N\in \mathcal{S}$ by construction; a normal $p^{\mathrm{\prime }}$-subgroup is by definition a member of $\mathcal{S}$. Hence $O_{p^{\mathrm{\prime }}}(G)$ is the largest normal $p^{\mathrm{\prime }}$-subgroup, and it is the only one with that property, since two normal $p^{\mathrm{\prime }}$-subgroups each contain the other.