Here [x] denotes the equivalence class of integers = x modulo n, where n will be specified according to context. We will sometimes write x instead of [x].
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1) This refers to the subgroup generated by 'powers' of 18, ie.
{[18]^n | n ∈ Z}
which can also be written as
{[n18] | n ∈ Z} = {[0], [18], [36], [54]},
since [18]^4 = [4x18] = [72] = [0].
2. Order 12, correct.
3. Order 12, correct.
4. Z_2 × Z_2, correct.
5. In order for an element ([x], y, [z]) ∈ Z_2 × Z × Z_4 to have finite order, we need y = 0. For if y ≠ 0, 'powers' of ([x], y, [z]) will never be equal to the (additive) identity of Z_2 × Z × Z_4 -- we can never get a ky = 0 for positive k.
So, all we need consider is elements of the form ([x], 0, [z]), and you should find that there are exactly eight of these:
(0, 0, 0), (0, 0, 1), (0, 0, 2) , (0, 0, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2) and (1, 0, 3).
Note that
{(0, 0, 0), (0, 0, 1), (0, 0, 2) , (0, 0, 3), (1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3)}
itself comprises a group that's isormorphic to Z_2 × Z_4 under the map ([x], 0, [y]) → ([x], [y]).
2007-03-17 22:41:33
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answer #1
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answered by MHW 5
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