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Unit 4: Lagrange's Theorem




          xH={xh | h  H} .                                                                     Notes
          Note that, if the group operation is +, then the right and left cosets of H in (G,+) represented by
          x  G are
          H + x = { h + x | h  H} and x + H = { x + h | h  H }, respectively.


                Example: Show that H is a right as well as a left coset of a subgroup H in a group G.
          Solution: Consider the right coset of H in G represented by e, the identity of G. Then
          He = { he | h  H } = ( h | h  H } = H .

          Similarly, eH = H.
          Thus, H is a right as well as left coset of H in G.


                Example: What are the right cosets of 4Z in Z?
          Solution: Now H = 4Z = { ......, -8, - 4, 0, 4, 8, 12, ..... }
          The right cosets of H are
          H + 0 = H, using Example.

          H+ 1 ={ ....., –11, –7, –3, 1, 5, 9, 13, .... )
          H + 2 = { ....., –10, –6, –2, 2, 6, 10, 14, .... )
          H + 3 = { ....., –9, –5, –1, 3, 7, 11, 15, .... )
          H + 4 = { ....., –8, –4, 0, 4, 8, 12 ,.....) = H

          Similarly, you can see that H + 5 = H + 1, H + 6 = H + 2, and so on.
          You can also check that H – 1 = H + 3, H – 2 = H + 2, H – 3 = H + l, and so on.
          Thus, the distinct right cosets are H, H + 1, H + 2 and H + 3.
          In general, the distinct right cosets of H (= nZ) in Z are H, H + 1, ....., H + (n – 1). Similarly, the
          distinct left cosets of H (= nZ) in Z are H, 1 + H, 2 + H, ....., (n – 1) + H.
          After understanding the concept of cosets. Let us discuss some basic and important properties of
          cosets.
          Theorem 1: Let H be a subgroup of a group G and let x, y  G.
          Then
          (a)  X  HX

          (b)  Hx = H  x  H.
          (c)  Hx = H xy   H.
                          –1
          Proof: (a) Since x = ex and e  H, we find that x  Hx.
          (b) Firstly, let us assume that Hx = H. Then, since x  Hx, x  H.

          Conversely, let us assume that x  H. We will show that Hx  H and H  Hx. Now any element
          of Hx is of the form hx, where h  H. This is in H, since h  H and x  H. Thus, Hx  H. Again, let
          h  H. Then h = (hx ) x  Hx, since hx   H.
                                        -1
                          -l
                H  HX.




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