That looks more like Z_2 * Z_4 ;)[tex]\mathbb{Z}_4\times\mathbb{Z}_2 = \{ (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3) \}[/tex]
AHA!! Geez I make some pretty stupid mistakes!The computation of <(2, 1)> is indeed faulty. You've written that (6, 3) = (2, 0), which is wrong. It should be (6, 3) = (2, 1), so that <(2, 1)> consists of (0, 0) and (2, 1).
Ive shown you the way that I do it - it is rather long and exhausting (particularly with more complicated examples), could you show me exactly how you would do these problems? (just one will suffice).And btw, you don't need to explicitly calculate the factor groups.
Yes I do see that G has order 8. But it is not easy for me, at this stage, to see that something like (1,1) has order 4, or (2,1) has order 2. THIS is what I lack. Some of my friends, and you, for instance, can just look at that and know its order...so how do you do it?The questions you've been given have simply been to find the order of the factor groups. It should seem obvious to you that for a finite group G, |G/H| = |G|/|H|. So in your case, you just need to find the order |H|. You're given the generator of H in your problems, like (1,1). It's not hard to see that this has order 4, and that G has order 4 x 2 = 8, so you know |G/H| = |G|/|H| = 8/4 = 2.
I don't know, maybe do a bunch of problems and it will start to come naturally two you. Well, I can give you the ideas but it's up to you to practice with them so they become natural. First, groups like [itex]\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \dots \times \mathbb{Z}_{n_k}[/itex] are abelian. Secondly, any element [itex](a_1, \dots , a_k)[/itex] is just [itex](a_1, 0, \dots , 0) + (0, a_2, 0, \dots , 0) + \dots + (0, \dots , 0, a_k )[/itex]. Finally, if you have an element [itex]x_1x_2\dots x_k[/itex], or, using additive notation, [itex]x_1 + x_2 + \dots + x_k[/itex], the order of this element is the least common multiple of the order of each of the [itex]x_i[/itex]. So, for example:Yes I do see that G has order 8. But it is not easy for me, at this stage, to see that something like (1,1) has order 4, or (2,1) has order 2. THIS is what I lack. Some of my friends, and you, for instance, can just look at that and know its order...so how do you do it?
Perfect, that is exactly what I needed!!!The order of (1,0) in is just the order of 1 in , and the order of (0,1) in is just the order of 1 in . Those numbers are, respectively, 2 and 4, whose least common multiple is 4. If we're looking at (1,2), then we have (1,2) = (1,0) + (0,2) and the order of (1,0) in is the order of 1 in , which is 2, and the order of 2 in is also 2. The least common multiple of 2 and 2 is 2.