SECTION 10 Cosets and the Theorem of Lagrange Theorem Let H be a subgroup of G. Let the relation  L be defined on G by a  L b if and only if a -1 b  H. Let the relation  R be defined on G by a  R b if and only if ab -1  H. Then  L and  R both equivalence relations on G Proof: exercise 26. The equivalence relation  L defines a partition of G, and the cell containing a is {ah | h  H}, which we denote by aH. Similarly for the equivalence relation  R.

Cosets Definition Let H be a subgroup of a group G. The subset aH = {ah | h  H} of G is the left coset of H containing a, while the subset Ha = {ha | h  H} is the right coset of H containing a.

Example Example: Exhibit the left cosets and the right cosets of the subgroup 3Z of Z. Solution: Our notation is additive, so the left coset of 3Z containing m is m+3Z. Let m=0, then 3Z = {…, -9, -6, -3, 0, 3, 6, 9, …} is one left coset. Then select 1 in Z but not in 3Z, and find the left coset containing 1. We have 1+3Z= {…, -8, -5, -2, 1, 4, 7, 10, …}. Observe 3Z and 1+3Z do not yet exhaust Z, then select 2 in Z but not in 3Z or 1+3Z, and find the left coset containing 2. We have 2+3Z= {…, -7, - 4, -1, 2, 5, 8, 11, …} Now 3Z, 1+3Z, and 2+3Z do exhaust Z, so these are all left cosest of 3Z. Since Z is abelian, the let coset m+3Z and the right coset 3Z+m are the same.

Example Fact: For a subgroup H of an abelian group G, the partition of G into left cosets of H and the partition into the right cosets are the same. In this case, we do not have to specify left or right cosets. Example: The group Z 6 abelian. Find the partition of Z 6 into cosets of the subgroup H={0, 3}. Solution: One coset is {0, 3} itself. Select 1 in Z 6 but not in {0, 3}. The coset containing 1 is 1+{0, 3} = {1, 4}. Similarly, select 2 in Z 6 but not in {0, 3} or {1, 4}, the coset containing 2 is 2+{0, 3} = {2, 5}. Since {0, 3}, {1, 4}, and {2, 5} exhaust all of Z 6, these are all cosets.

Example Let H be the subgroup ={  0,  1 } of S 3. Find the partitions of S 3 into left cosets of H, and the partition into right cosets of H. (See the table for the symmetric group S 3 on the next slide.) Solution: For the partition into left cosets, we have H={  0,  1 }  1 H={  1  0,  1  1 } ={  1,  3 }.  2 H={  2  0,  2  1 } ={  2,  2 }. The partition into right cosets is H={  0,  1 } H  1 ={  0  1,  1  1 } ={  1,  2 }. H  2 ={  0  2,  1  2 } ={  2,  3 }. Since S 3 is not abelian, the partition into left cosets of H is different from the partition into right cosets.

The Theorem of Lagrange Fact: Every coset (left or right) of a subgroup H of a group G has the same number of elements as H. Theorem of Lagrange Let H be a subgroup of a finite group G. Then the order of H is a divisor of the order of G. Proof: Let n be the order of G, and let H have order m. The fact above shows every coset of H also has m elements. Let r be the number of cells in the partition of G into left cosets of H. Then n=rm, so m is indeed a divisor of n.

More Theorem Corollary Every group of prime order is cyclic. Proof: exercise. Theorem The order of an element of a finite group divides the order of the group. Proof: it follows directly from the Theorem of Lagrange.

Index Definition Let H be a subgroup of a group G. The number of left cosets of H in G is the index (G: H) of H in G. Note: if G is finite, then (G: H) is finite and (G: H)=|G| / |H|, since every coset of H contains |H| elements. Theorem Suppose H and K are subgroups of a group G such that K  H  G, and suppose (H : K) and (G: H) are both finite. Then (G: K) is finite, and (G: K)= (G: H) (H : K)