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Fill in the blanks.

1) The cyclic subgroup of Z_24 generated by 18 has order__.
2) Z_3 × Z_4 is of order __.
3) The element (4,2) of Z_12 × Z_8 has order __.
4) The Klein 4-group is isomorphic to Z__ × Z__.
5) Z_2 × Z × Z_4 has __ elements of finite order.

I solved some of them but I'm not so sure about the answers.
So, can someone check my answers?
(1) I don't get this. Please help me with this!!
(2) I got order of 12.
(3) I got order of 12.
(4) I got Z_2 × Z_2.
(5) I'm not sure about this. Please help me.

So, can you check my answers and help me with the problems
that I wasn't able to do? Thank you.

2007-03-17 20:27:00 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

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].

***

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 · answer #1 · answered by MHW 5 · 0 0

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