Quote:
Originally Posted by smcdonn2
2.
0= {0}=order1
1= {0,1,2,3}=4 elements
2={0,2} two elements
3={0,3,2,1}= 4elements
3.
{(0,0),(0,2),(2,0),(2,2)}
So let me ask, with something like Z/2Z X Z/2Z. Do i work the set out first for each individual then take the ordered operation? So number 3 above would be the correct way to compute it. Or should I start with the identity, in this case (0,0) i hope, and add the generator (1,1) then perform the given mod op? So i was thinking the way to do Z_2 X Z_3:
Take the identity and add the generator
{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2),}
And in all honestly Im not sure what Z/4Z, It looks like its just another way to write Z mod 4. which is what I read. But something more complex like:
(3,1)+<(1,1)> in (z_4Xz_4)/<(1,1)>
i have no clue what this says
First of all, mathematics is a very precise thing. You should read your statements before you send them, so that your question is clear. This also gets you in the habit of writing well, and whoever reads your proofs will appreciate that.
Z/4Z is the factor group (or quotient group, depending on the language your book uses) that you're calling "Z mod 4". I would call it "Z mod 4Z", and I write it Z/4Z instead of Z_4, because (at least for p prime), Z_p means something else to a lot of mathematicians.
It is the set of cosets of the subgroup 4Z in Z. There are 4 such cosets:
0+4Z, 1+4Z, 2+4Z, and 3+4Z. This forms a group under the operation
(a+4Z) + (b+4Z) = (a+b)+4Z.
We often omit the (implied) 4Z when talking about this group and list the elements as 0,1,2,3.
To your questions above:
Z/2Z x Z/2Z has 4 elements: {(0,0), (0,1), (1,0), (1,1)}.
Your calculation that (1,1) has order 6 in Z/2Z x Z/3Z is correct.
Quote:
(3,1)+<(1,1)> in (z_4Xz_4)/<(1,1)>
First, <(1,1)> is a (normal) subgroup of Z/4Z x Z/4Z. (Z/4Z x Z/4Z)/<(1,1)> is the set of cosets of <(1,1)>.
<(1,1)> = {(0,0), (1,1), (2,2), (3,3)}
(3,1)+<(1,1)> = {(3,1), (0,2), (1,3), (2,0)}
Notice that (3,1)+<(1,1)> = (2,0)+<(1,1)>.
(3,1) and (2,0) are different choices of
coset representatives for this coset.
In other words, (3,1) and (2,0) are the same element in (Z/4Z x Z/4Z) / <1,1> in just the same way as 1, 5, and 9 are the same element in Z/4Z.
Lastly, 2 elements represent the same coset if and only if their difference is in the normal subgroup (in this case, notice (3,1)-(2,0) = (1,1), which is in <(1,1)>).
edit: You can now see why I write Z/4Z. This group is Z/<4>, and <4> = 4Z.
Maybe this is helpful and will spark some questions.