232 CHAPTER 4 Abstract Algebra
reflexivity: g≡ g( modH) sinceg−^1 g=e∈H.
symmetry: If g ≡ g′( modH) then g−^1 g′ ∈ H, and so g′−^1 g =
(g−^1 g′)−^1 ∈H. Therefore, alsog′≡ g( modH)
transitivity: Ifg ≡ g′( modH) andg′ ≡ g′′( modH), then g−^1 g′,
g′−^1 g′′∈H. But theng−^1 g′′=g−^1 g′g′−^1 g′′ ∈H, proving that also
g≡ g′′( modH).Pretty easy, eh?
As a result we see thatGis partitioned into mutually disjoint equiva-
lence classes. Next we shall actually determine what these equivalence
classes look like. Thus let g∈ Gand let [g] be the equivalence class
(relative to the above equivalence relation) containingg.
We Claim: [g] =gH={gh|h∈H}.
Proof of Claim: Note first that an element ofgH looks likegh, for
someh ∈H. Sinceg−^1 (gh) = h ∈H we see that g ≡gh( modH),
i.e., gh ∈ [g]. This proves that gH ⊆ [g]. Conversely, assume that
g≡ g′( modH), i.e., thatg≡ g′( modH), which says thatg−^1 g′∈H.
But theng−^1 g′=hfor someg′=gh∈gH. This proves that [g]⊆gH
and so [g] =gH.
Next we would like to show thatHis a finite subgroup ofGthen the
elements of each equivalence classgH, g∈Ghave the same number of
elements. In fact, we shall show that|gH|=|H|, for eachg∈G. To
prove this we shall define a mappingf :H→gH and show that it is
a bijection. Namely, we definef(h) =gh, h∈H.
fis one-to-one: Ifh, h′∈Hand iff(h) =f(h′), thengh=gh′. We
now multiply each side byg−^1 and get h=g−^1 gh=g−^1 gh′=h′.
Thusf is one-to-one.
f is onto: Ifgh∈gH, thengh=f(h) and sof is onto.It follows, therefore, that |gH| = |H| for each element g ∈G. If G
is also a finite group, this says thatGis partitioned into sets, each of