4.6 Cosets and Lagrange’s Theorem
Definition 4.13 Let \((G, *)\) be a group, H a subgroup of G, and \(g \in G\).
- The left coset of H by g is \(gH := \{ g*h : h \in H\}\).
- The right coset of H by g is \(Hg := \{ h*g : h \in H\}\).
We write \(G:H\) for the set of left cosets of H by elements of G so \(G:H = \{ gH : g \in G\}\), and \(|G:H|\) for its size. Similarly \(H:G\) is the set of right cosets of H by elements of G.
Let’s look at some example cosets.
Example 4.8
- The 3-cycle \((1,2,3) \in S_4\) has order 3, so \(H=\langle (1,2,3) \rangle\) is equal to \(\{ e, (1,2,3), (1,2,3)^2 = (1,3,2)\}\). Then \[\begin{align*} (1,2)H &= \{ (1,2), (1,2)(1,2,3), (1,2)(1,3,2) \} \\ &= \{ (1,2), (2,3), (1,3) \} \\ H(1,2) &= \{ (1,2), (1,2,3)(1,2), (1,3,2)(1,2) \} \\ &= \{ (1,2), (1,3), (2,3) \}. \end{align*}\] So in this case, \((1,2)H = H(1,2)\).
- The 2-cycle \((1,2) \in S_3\) has order 2, so \(H := \langle (1,2) \rangle\) is equal to \(\{ e, (1,2) \}\). Then \[\begin{align*} (2,3)H &= \{ (2,3), (2,3)(1,2)\} = \{ (2,3),(1,3,2)\} \\ H(2,3)&= \{ (2,3), (1,2)(2,3) \} = \{ (2,3), (1,2,3) \} \end{align*}\] So in this case \((2,3)H \neq H(2,3)\).
- We know that \(2\ZZ = \{ 2 z : z \in \ZZ\}\), the set of all even integers, is a subgroup of \((\ZZ,+)\). Then the coset \(1+2\ZZ = \{ 1+ 2z : z \in \ZZ\}\) is the set of all odd integers. More generally if \(n \in \ZZ\) then \(n\ZZ\) is a subgroup of \(\ZZ,+\) and \(1 + n\ZZ\) is the set of all integers congruent to 1 mod n, \(2+n\ZZ\) is the set of all integers congruent to 2 mod n, and so on.
Consider the cosets \(0+3\ZZ=3\ZZ, 1+3\ZZ, 2+3\ZZ\) of the subgroup \(3\ZZ\) of \((\ZZ,+)\). Each element of \(\ZZ\) belongs to exactly one of them, for any integer is congruent to exactly one of 0, 1, or 2 mod 3. So we have \[\begin{equation*} \ZZ = 3\ZZ \sqcup (1+3\ZZ) \sqcup (2+3\ZZ) \end{equation*}\] (the symbol \(\sqcup\) means that this is a disjoint union). Furthermore, \(3n \mapsto i+3n\) is a bijection between \(3\ZZ\) and \(i+3\ZZ\), so all three cosets have the same size. So the cosets of the subgroup \(3\ZZ\) split \(\ZZ\) into three disjoint pieces, each of the same size. If \(\ZZ\) were a finite set this would imply that its size was three times that of the subgroup \(3\ZZ\). When we prove Lagrange’s theorem, which says that if G is finite and H is a subgroup then the order of H divides that of G, our strategy will be to prove that you get exactly this kind of decomposition of G into a disjoint union of cosets of H.
This last example is important, because it tells us we can have \(g_1H = g_2 H\) even if \(g_1 \neq g_2\). The next result tells us exactly when this happens.
From now on I’ll be lazy and write \(gh\) instead of \(g*h\).
Proof.
- (R) Let \(g\in G\). Then \(g^{-1}g=e \in H\) as H is a subgroup, so \(g \sim g\).
- (S) Let \(g \sim k\). Then \(g^{-1}k \in H\), so \((g^{-1}k)^{-1} \in H\) as H is a subgroup. But \((g^{-1}k)^{-1} = k^{-1}( g^{-1})^{-1}= k^{-1}g\), so \(k \sim g\).
- (T) Let \(g\sim k \sim l\). Then \(g^{-1}k \in H, k^{-1}l\in H\), so \((g^{-1}k)(k^{-1}l) \in H\) as H is a subgroup. But \(g^{-1}kk^{-1}l=g^{-1}l\), so \(g \sim l\).
Of course, there’s a version of this for right cosets as well
Each coset of a subgroup H has the same size as H.
Proof. We will prove the first of these equalities, the second being similar. The proof is by writing down a bijection between H and \(gH\). Define \(\alpha : H \to gH\) by \(\alpha(h) = g*h\).
\(\alpha\) is one-to-one: for if \(\alpha(h_1)=\alpha(h_2)\) then \(g*h_1=g*h_2\), so multiplying on the left by \(g^{-1}\) we get that \(h_1=h_2\).
\(\alpha\) is onto: for any element of \(gH\) equals \(g*h\) for some \(h \in H\), and \(\alpha(h)=g*h\), so \(g*h\) is in the image of \(\alpha\).
Thus \(\alpha\) is a bijection, and \(|gH| = |H|\).Theorem 4.2 (Lagrange’s Theorem) Let G be a finite group and let H be a subgroup. Then \(|G| = |G:H||H| = |H:G||H|\). In particular, the order of H divides the order of G.