How do you find the order of an element in a factor group?

Question

For example:
the order of 5+<4> in the factor group Z(sub 12)/<4>
or (2,1)+<(1,1)> in (Z(sub3)xZ(sub6))/<(1,1)>

Answer 1

For 5 + <4>. Keep adding the quantity to itself until you get the identity <4> = {4,8,12}
So the next element is 10 + <4> and then 3 + <4> and then
8 + <4> = <4>. So the order of the element is 4
In The next one write out the elements of <1,1> (excuse shorthand) you get { 1,1 2,2 0,3 1,4 2,5 0,0}.
Now add 2,1 + <1,1> to itself as we did in the first example until the k2, k1 is in <1,1>, then k is the order.
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Answer 2

hi, permit n be the order. Then n is the smallest answer to (3n,3n) = (0,0) mod (a million,2) so there is a few integer x with (3n,3n) = (x,2x) in (Z4 x Z8). this provides the gadget 3n = x mod 4 and 3n = 2x mod 8, so (2x mod 8) mod 4 = 2x mod 4 = x mod 4. 2x = x mod 4 has the answer x = 0 mod 4. as a consequence 2x = 0 mod 8, so 3n = 0 mod 8 and n = 0 mod 8. The smallest such answer is the order n = 8.

Answer 3

Your notation isn't familiar to me, but wouldn't 5 be the same thing as 1? If so, its order is obviously 1. If not, then when you square the thing you get 25, which is a lot like 1, so its order is 2.

In the second case, if you square the thing you get the identity, so its order is 2.