Group Theory/Cosets and Lagrange's theorem

left coset Let be a group, and a subgroup, and . Then the left coset of represented by is the set . Right cosets are defined in an analogous fashion: right coset Let be a group, and a subgroup, and . Then the right coset of represented by is the set . For both of these, we have the following proposition: being in the same left coset is an equivalence relation Let be a group, and define a relation on by . Then is an equivalence relation, and we also have the formula . We first prove the alternative formula for being in the same coset. If and , we find from the latter equation an so that , so that and hence , the latter identity because is a group and in particular closed under inversion. On the other hand, if , then for some , and hence (because the identity is in ) and then . Hence, the two formulae for the relations coincide, and it remains to check that we're dealing with an equivalence relation. Indeed, suppose and . Then there exist so that and so that . Then , so that , ie. , proving transitivity. Reflexivity follows since the identity is in , and symmetry follows because implies and hence . Analogously, we have the following proposition: being in the same right coset is an equivalence relation Let be a group, and define a relation on by . Then is an equivalence relation, and we also have the formula . Consider the opposite group of . There is a bijection , given by , under which two elements belonging to the same right coset of correspond to elements belonging to the same left coset of . But the relation defined by the latter was seen to be an equivalence relation. index Let be a group, and let be a subgroup. Then the index of is defined to be the number . That is, the index is precisely the number of left cosets. Lagrange's theorem Let be a group, and let be a finite subgroup. Then . In particular, the order of divides the order of . We have seen that being in the same left coset is an equivalence relation, so that the equivalence classes partition . Moreover, every equivalence class (ie. coset) has the same cardinality as via the bijection . number of right cosets equals number of left cosets Let be a group, and a subgroup. Then the number of right cosets of equals the number of left cosets of . By Lagrange's theorem, the number of left cosets equals . But we may consider the opposite group of . Its left cosets are almost exactly the right cosets of ; only the orders of the products are interchanged. But in particular, the number of left cosets of , which, by Lagrange's theorem equals , is equal to , which is what we wanted to prove. Hence, we may also use the notation for the number of right cosets. degree formula Let be a group, and let . Then . We may partition into a family of left cosets , where for all we have . Moreover, may be partitioned into a family of left cosets of . Then is a family of left cosets of that partitions (since each element is in one left coset of , and then is in a unique coset , and then is the unique coset in which is), and the cardinality of this family, which is , is the number of left cosets of in . Exercises Prove that , thus establishing another formula for the equivalence relation of being in the same coset. Formulate Lagrange's theorem for right cosets, without using index notation.