Solution for Section G.2 Question 1a

1a. Check the three subgroup conditions:

• identity: is the identity element of G in H?
  [0] is the identity element of the group G. [0] is also in H.
  
• inverse: for each element a in H, is a^-1 also in H?
  In the group G, [a]^-1 = [10-a]. So
  [0]^-1 = [0], which is in H.
  [2]^-1 = [8], which is in H.
  [4]^-1 = [6], which is in H.
  [6]^-1 = [4], which is in H.
  [8]^-1 = [2], which is in H.
  Hence each element in H has its inverse in H.
  
• closure under addition modulo 10: if you add any pair of elements in H, is the result also an element of H?
  We need to check that for all a, b in H, a + b is in H, where + is addition modulo 10.
  [0] + [0] = [0], [0] + [2] = [2], [0] + [4] = [4], [0] + [6] = [6], [0] + [8] = [8].
  [2] + [0] = [2], [2] + [2] = [4], [2] + [4] = [6], [2] + [6] = [6], [2] + [8] = [0].
  [4] + [0] = [4], [4] + [2] = [6], [4] + [4] = [8], [4] + [6] = [0], [4] + [8] = [2].
  [6] + [0] = [6], [6] + [2] = [8], [6] + [4] = [0], [6] + [6] = [2], [6] + [8] = [4].
  [8] + [0] = [8], [8] + [2] = [0], [8] + [4] = [2], [8] + [6] = [4], [8] + [8] = [6].
  Hence H is closed.

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