Normal Subgroups and Quotient Groups

Condition 1 implies condition 2 because equality implies containment.

Suppose condition 2 holds. Applying it to $g^{-1}$ gives $g^{-1}Ng\subseteq N$; conjugating this containment by $g$ and using $(g^{-1})^{-1}=g$ gives $N\subseteq gNg^{-1}$, while condition 2 gives the reverse containment. Hence $gNg^{-1}=N$ for every $g$, so condition 1 holds.

Suppose condition 1 holds. If $x\in gN$, then $x=gn=(gng^{-1})g$ for some $n\in N$, and $gng^{-1}\in N$ by [F1], so $x\in Ng$. Replacing $g$ by $g^{-1}$ gives the reverse inclusion, hence $gN=Ng$.

Suppose condition 3 holds. For $n\in N$, the element $gn$ lies in $gN=Ng$, so $gn=n'g$ for some $n'\in N$; therefore $gng^{-1}=n'\in N$. Thus $gNg^{-1}\subseteq N$ and condition 2 holds.

Steps 1.1 through 1.4 prove that conditions 1, 2, and 3 are equivalent.

For every $g\in G$, commutativity gives $gH=\{gh:h\in H\}=\{hg:h\in H\}=Hg$.

Hence $H\mathrel{\trianglelefteq}G$ by the coset characterisation of normality.

If $g\notin H$, then $gH\ne H$ by [L3]. Since [F1] and the hypothesis give exactly two left cosets, [L1] shows that $H$ and $gH$ are disjoint and cover $G$, so $gH=G\setminus H$.

By [L2] there are exactly two right cosets. If $g\notin H$, then $Hg\ne H$ by [L3]; the same elementary coset argument shows that distinct right cosets are disjoint and cover $G$, so $Hg=G\setminus H$.

If $g\in H$, then $gH=H=Hg$ by [L3]; if $g\notin H$, steps 1.1 and 1.2 give $gH=G\setminus H=Hg$. Thus $gH=Hg$ for every $g\in G$, and [L4] gives $H\mathrel{\trianglelefteq}G$.

By [L1], the set $K=\bigcap_{N\in\mathcal N}N$ is a subgroup of $G$.

Fix $g\in G$ and $x\in K$. For every $N\in\mathcal N$, one has $x\in N$ and $N\mathrel{\trianglelefteq}G$, so $gxg^{-1}\in N$ by [L2]. Hence $gxg^{-1}\in K$.

Thus $gKg^{-1}\subseteq K$ for every $g\in G$, and [L2] gives $K\mathrel{\trianglelefteq}G$.

The identity lies in $Z(G)$. If $x,y\in Z(G)$ and $g\in G$, then $(xy)g=x(yg)=x(gy)=(xg)y=(gx)y=g(xy)$, so $xy\in Z(G)$. If $x\in Z(G)$, then $xg=gx$ implies $x^{-1}g=gx^{-1}$ after multiplying by $x^{-1}$ on both sides, so $x^{-1}\in Z(G)$. Hence $Z(G)\le G$.

If $z\in Z(G)$ and $g\in G$, then $gzg^{-1}=zgg^{-1}=z$. Therefore $gZ(G)g^{-1}=Z(G)$ for every $g\in G$.

Steps 1.1 and 1.2 show that $Z(G)$ is a subgroup invariant under conjugation, so it is normal.

Direct multiplication gives $x[g,h]x^{-1}=[xgx^{-1},xhx^{-1}]$ for all $g,h\in G$.

The conjugate $x^{-1}Dx=\{x^{-1}dx:d\in D\}$ is a subgroup of $G$.

For every commutator $c=[g,h]$, step 1.1 gives $xcx^{-1}\in D$, so $c\in x^{-1}Dx$. Thus the subgroup $x^{-1}Dx$ contains every generator of $D$, and [L1] gives $D\subseteq x^{-1}Dx$.

Conjugating the containment in step 2.1 by $x$ gives $xDx^{-1}\subseteq D$. Since $x$ was arbitrary, [L2] gives $D\mathrel{\trianglelefteq}G$.

Suppose $H\mathrel{\trianglelefteq}G$ and $aH=a'H$, $bH=b'H$. By [L2], write $a'=ah_1$ and $b'=bh_2$ with $h_1,h_2\in H$. Then $(ab)^{-1}a'b'=b^{-1}h_1bh_2\in H$ by [L1] and [F2], so [L2] gives $abH=a'b'H$. Hence [F1] is independent of both representatives.

Conversely, suppose [F1] is well defined. For $h\in H$ and $g\in G$, the equal cosets $H=eH=hH$ give the same product with $gH$, so $gH=(eH)(gH)=(hH)(gH)=hgH$.

The equality $gH=hgH$ gives $g^{-1}hg\in H$ by [L2]. Thus $g^{-1}Hg\subseteq H$ for every $g$, and [L1] gives $H\mathrel{\trianglelefteq}G$.

Step 1.1 proves sufficiency and steps 1.2 and 2.1 prove necessity, establishing the biconditional.

By [L1], the formula in [F1] is a binary operation on the coset set, independent of representatives.

For $g,h,k\in G$, one has $((gN)(hN))(kN)=(gh)kN=g(hk)N=(gN)((hN)(kN))$. Also $(eN)(gN)=gN=(gN)(eN)$, so $N=eN$ is the identity.

The products $(gN)(g^{-1}N)$ and $(g^{-1}N)(gN)$ both equal $eN=N$, so $g^{-1}N$ is the inverse of $gN$.

Steps 1.1 through 1.3 verify the binary operation, associativity, identity, and inverse axioms; therefore $G/N$ is a group with the stated identity and inverses.

For $g,h\in G$, one has $\pi(gh)=ghN=(gN)(hN)=\pi(g)\pi(h)$, so $\pi$ is a group homomorphism.

Every element of $G/N$ is a coset $gN=\pi(g)$ for some $g\in G$, so $\pi$ is surjective.

Hence the canonical projection is a surjective group homomorphism.

If $[G:N]$ is finite, then by [F1] the coset set underlying $G/N$ is finite with cardinality $[G:N]$; hence [F2] and [L2] give $|G/N|=[G:N]$.

If $G$ is finite, then [L1] gives $|G|=[G:N]|N|$. Since $N$ contains the identity, $|N|\ne0$, and step 1.1 yields $|G/N|=[G:N]=|G|/|N|$.

The two asserted formulas follow.

Suppose $G/N$ is abelian. For $g,h\in G$, the commutator of the cosets $gN$ and $hN$ is the identity, so [L1] gives $[g,h]N=N$; hence $[g,h]\in N$ by [L3].

Conversely, suppose $[G,G]\subseteq N$. Then for any $g,h\in G$, one has $[g,h]\in N$, so [L3] and [L1] show that the commutator of $gN$ and $hN$ is $N$. Multiplying the equality $(gN)(hN)(gN)^{-1}(hN)^{-1}=N$ on the right by $(hN)(gN)$ gives $(gN)(hN)=(hN)(gN)$. Thus $G/N$ is abelian.

The subgroup $N$ contains every commutator, so it contains the subgroup they generate: $[G,G]\subseteq N$.

Steps 1.1 and 2.1 prove the forward implication, and step 1.2 proves the reverse implication.

For arbitrary cosets $gN,hN\in G/N$, commutativity in $G$ gives $(gN)(hN)=ghN=hgN=(hN)(gN)$.

Hence every two elements of $G/N$ commute, so $G/N$ is abelian.

The set $n\mathbb Z$ contains $0=n0$; if $nk,n\ell\in n\mathbb Z$, then $nk+n\ell=n(k+\ell)$ and $-(nk)=n(-k)$ also lie in $n\mathbb Z$. Hence $n\mathbb Z\le(\mathbb Z,+)$.

For $a,x\in\mathbb Z$, one has $x\in a+n\mathbb Z$ if and only if $x=a+nq$ for some $q\in\mathbb Z$, if and only if $x\equiv a\pmod n$, if and only if $x\in[a]_n$. Therefore $a+n\mathbb Z=[a]_n$.

Since $(\mathbb Z,+)$ is abelian, the subgroup $n\mathbb Z$ is normal.

Under the equality in step 1.2, [L3] and [F4] give $(a+n\mathbb Z)+(b+n\mathbb Z)=(a+b)+n\mathbb Z=[a+b]_n=[a]_n+[b]_n$.

Steps 2.1, 1.2, and 2.2 show that the quotient group and the group of congruence classes have the same underlying set and operation; [L4] confirms the published group convention, including $n=0$ and $n=1$.