G4 - applications of factor groups

the center of a group is the set of all elements that commute with every other element of the group, ie. $Z(G)=\{a\in G\mid ax=xa\mathrm{\forall }x\in G\}$

if $G$ is abelian, all elements commute, ie. $Z(G)=G$, so it suffices to show that $G/Z(G)$ has only the identity coset $Z(G)$

let $G/Z(G)=⟨gZ(G)⟩$, and $a\in G$, then $\mathrm{\exists }i\in \mathbb{Z}$ such that $aZ(G)=(gZ(G){)}^i=g^{i}Z(G)$ $⟹a=g^{i}z$ for some $z\in Z(G)$

since $g^{i}z\in C(g)$, so does $a$

as $a$ is an arbitrary element of $G$, this means that every element of $G$ commutes with $g$, so $g\in Z(G)$

thus, $gZ(G)=Z(G)$ is the only element of $G/Z(G)$

this theorem shows that if $G/H$ is cyclic for $H\le Z(G)$, then $G$ is abelian

the contrapositive of the theorem is most often used, ie. if $G$ is non-abelian, then $G/Z(G)$ is not cyclic

eg: from this and lagrange's theorem, a non-abelian group of order $pq$ must have a trivial center, where $p$ and $q$ are primes

if $G/Z(G)$ is cyclic, it must be trivial

eg: let $H⊲G$ and $\overline{K}=\{H,a_{1}H,a_{2}H,a_{3}H\}\le G/H$

the pull back of $\overline{K}$ to $G$, $K=H\cup a_{1}H\cup a_{2}H\cup a_{3}H$ is a subgroup of $G$ of order $4|K|$

let $a_0=e$ and $a_i,a_j\in \{a_0,a_1,a_2,a_3\}$, then for any $a_i{h}_i,a_j{h}_j\in K$, $(a_i{h}_i)(a_j{h}_j)\in a_{i}H{a}_{j}H=a_{k}H$ for some $k=0,1,2,3$, so $(a_i{h}_i)(a_j{h}_j)\in K$

eg: suppose the factor group $G/H$ of a finite group $G$ has an element $aH$ of order $n$, then $G$ has an element of order $n$

let $|a|=k$, then in $G/H$, $(aH{)}^k=a^{k}H=H$

$k$ must be a multiple of $n$, so $k=mn$, and $(a^m{)}^n=a^k=e$, and $n$ is the smallest such positive integer