I will use this result as part of Kummer's proof on Fermat's Last Theorem for regular primes.

Definition 1: Normal Subgroup

A subgroup H of a group G is called a normal subgroup of G if for all a ∈ G and all h ∈ H, there exists an element h' ∈ H such that ah = (h')a

NOTE: It is quite possible that h'=h but this is not necessary. From this perspective, normal subgroup is not necessarily an abelian group, that is, it may not be commutative (see here for definition of an abelian group).

NOTE #2: For all purposes definition 1 is the same as aH = Ha. Here's why:

For any x ∈ aH, there exists h ∈ H such that x = ah. From the definition, there exists h' such that h' ∈ H and ah=h'a which means that x=h'a which means that x ∈ Ha.

Lemma 1: A subgroup H is normal in G if and only if g

^{-1}Hg ⊆ H for all g in G.

Proof:

(1) Assume H is a normal subgroup of G.

(2) Then, for any g ∈ G, h ∈ H, there exists h' ∈ H such that hg = gh' [See Definition of a Normal Subgroup above]

(3) Thus, g

^{-1}hg = h' [Since g

^{-1}gh' = eh' = h']

NOTE: We know that g

^{-1}∈ G and that g

^{-1}g = e (Inverse Property of Groups, see here). We know that eh' = h' (Identity Property of Groups, see here)

(4) Therefore for all g,h, we know that g

^{-1}hg ∈ H [Since g

^{-1}hg = h' and h' ∈ H]

(5) So that g

^{-1}Hg ⊆ H [That is, for all x ∈ g

^{-1}Hg → x ∈ H]

(6) Assume that g

^{-1}Hg ⊆ H for all g ∈ G

(7) This means that Hg ⊆ gH [Since from #6, we have gg

^{-1}Hg ⊆ gH and gg

^{-1}Hg = eHg = Hg]

(8) Since g

^{-1}∈ G, our assumption in #6 gives us:

(g

^{-1})

^{-1}H(g

^{-1}) = gHg

^{-1}⊆ H

But this also means that:

gH ⊆ Hg since gHg

^{-1}g = gH ⊆ Hg

(9) So combining step #7 and step #8 gives us: Hg = gH

NOTE: Hg = gH shows that for all h ∈ H, g ∈ G, there exists h' ∈ H such that: hg = gh'.

QED

Lemma 2: A subgroup of an abelian group is normal

Proof:

(1) Let G be an abelian group. [See here for definition of an abelian group]

(2) Let H be a subgroup of G. [See here for definition of a subgroup]

(3) g ∈ G, h ∈ H → gh = hg. [from the definition of abelian groups]

(4) Multiplying both sides by the inverse of g (see here) gives us:

g

^{-1}gh = g

^{-1}hg

This implies that:

h = g

^{-1}hg

(5) Since h ∈ H, this gives us:

h ∈ H, g ∈ G → g

^{-1}hg ∈ H.

(6) Using Lemma 1 above, we can now conclude that H is a normal subgroup of G.

QED

Definition 2: Quotient Group

Let G be a group and H a normal subgroup of G. The Quotient Group is defined as as the coset aH such that {aH : a ∈ G} and is represented as G/H

Example 1: Z/2Z under addition = { 0+2Z, 1+2Z}

Let Z be the set of integers.

We can see that the left coset 2z = the set of even integers = { ..., -2, 0, 2 ... }

Z/2Z = {a(2Z) : a ∈ Z} = two distinct cosets {0 + 2z, 1 + 2z}

Lemma 3: if aH = a'H, then there exists h ∈ H such that a' = ah

Proof:

(1) a' ∈ a'H [H is a group → e ∈ H → a'e ∈ a'H]

(2) a' ∈ aH [since aH = a'H]

(3) From #2, there exists h ∈ H such that a' = ah

QED

Lemma 4: aH = H if and only if a ∈ H

Proof:

(1) Assume aH = H

(2) a = ae ∈ aH

(3) So combining #1 and #2, a ∈ H

(4) Assume a ∈ H

(5) x ∈ aH → x=ah with h ∈ H

(6) But if a ∈ H and h ∈ H, then ah ∈ H (by property of closure, see here)

(7) From #5 and #6, we conclude that aH ⊆ H

(8) Let h ∈ H

(9) From #4 and #8, ah ∈ H and a

^{-1}h ∈ H

(10) So, h = eh = (aa

^{-1})h = a(a

^{-1}h)

(11) So, from #10, h ∈ aH and therefore H ⊆ aH

(12) Combining #7 (aH ⊆ H) and #11 (H ⊆ aH) gives us: aH = H

QED

Lemma 5: A Quotient Group is itself a group under the operation (aH)(bH) = abH.

Proof:

(1) Let H be a normal subgroup of G.

(2) Let G/H be the quotient group { aH : a ∈ G }

(3) First, we need to prove that (aH)(bH) = abH is a well-defined operation. In other words, we need to show that aH=a'H, bH = b'H → (a'b')H = (ab)H

NOTE: A function maps each of its input to exactly one output (see here for Wikipedia's review of a mathematical function) So, we need to prove if the inputs are equal, the outputs are equal.

(3) Assume that aH = a'H and bH = b'H with a,a',b,b' ∈ G

(4) Then a' = ah

_{1}and b' = bh

_{2}for some h

_{1},h

_{2}in H [By Lemma 3 above]

(5) Then a'b'H = (ah

_{1})(bh

_{2})H

(6) Using Lemma 4 above, since h

_{1}, h

_{2}∈ H we know that:

h

_{1}H = H

h

_{2}H = H

So that we have:

(ah

_{1})(bh

_{2})H = (ah

_{1})bH

(7) Since H is a normal subgroup of G, we know that b ∈ G → bH=Hb [See definition 1 above, note 2]

This gives us:

(ah

_{1})bH = (ah

_{1})Hb

Using Lemma 4 above gives us

(ah

_{1})Hb = aHb = abH

(8) Closure: x ∈ G/H, y ∈ G/H → xy ∈ G/H since:

x ∈ G/H → there exists a ∈ G

_{}such that x = aH.

y ∈ G/H → there exists b ∈ G

_{}such that y = bH

_{}

xy = (aH)(bH) = (ab)H [Definition of operation]

xy ∈ G/H since a ∈ G, b ∈ G → ab ∈ G so (ab)H ∈ G/H.

(9) Identity: H

e ∈ G → eH ∈ G/H

e ∈ H so eH = H [From Lemma 4 above]

Further, we see that:

(aH)(H) = (aH)(eH) = (ae)H = aH

(14) Inverse: a

^{-1}H

x ∈ G/H → x = aH such that a ∈ G

a ∈ G → a

^{-1}∈ G

So a

^{-1}H ∈ G/H

Further,

(aH)(a

^{-1}H) = (aa

^{-1})H = eH = H

(15) Associativity:

(aHbH)cH = (ab)HcH = (ab)cH [From the definition of the operation for this group]

= a(bc)H = [(ab)c = a(bc) since a,b,c ∈ G and G is a group so is characterized by associativity]

= aH(bc)H = [From the definition of operation for this group: aH(bc)H = a(bc)H]

= aH(bHcH) [From the defintion of operation for this group: bHcH = (bc)H]

QED