1. ## Cylic

1. Suppose G = Z12 x Z12 and H =<(2,2)> is the subgroup generated by (2,2). Find the order of the element (5, 8) + H in G/H and prove it. Is G/H cyclic? Explain.

2. Find a direct product of cyclic groups of prime power order that is isomorphic to (Z9 x Z9)/<(3,3)>, prove your answer.

2. ## Re: Cylic

1. this is just straight-forward calculation:

H = {(2,2), (4,4), (6,6), (8,8), (10,10), (0,0)}. this has 6 elements, so so does (5,8) + H:

{(7,10), (9,0), (11,2), (1,4), (3,6), (5,8)}.

calculating powers of (5,8) + H, we get:

2[(5,8) + H] = (10,4) + H
3[(5,8) + H] = (3,0) + H
4[(5,8) + H] = (8,8) + H = H, so (5,8) + H is of order 4.

for G/H to be cyclic, it would need to have an element of order 144/6 = 24.

but G/H has no element of order > 12, since 12[(a,b) + H] = (12a, 12b) + H = (0,0) + H = H.

2. what have you tried so far?

3. ## Re: Cylic

Thank you so much for the first one.

And for the second one
I'm kinda clueless

I know <(3, 3)> is the cylic subgroup generated by (3, 3)

Z is of order 81
What do i do about the prime power order?

4. ## Re: Cylic

well <(3,3)> has order 3 right? so Z9xZ9/<(3,3)> has order 27. no matter how you slice and dice it, it looks like you'll have 2 choices:

1. Z9xZ9/<(3,3)> ≅ Z3 x Z9
2. Z9xZ9/<(3,3)> ≅ Z3 x (Z3 x Z3).

how can you tell which one you have? well, if you have an element of order > 3, you DON'T have the second one.

so, tell me, what is the order of (0,1) + <(3,3)>?

4

is it?

7. ## Re: Cylic

you sound unconvinced, as well you should. 4 does not divide 27, so cannot be the order of any element of Z9xZ9/<(3,3)>, which has order 27. why don't you actually see what is the smallest m such that (0,m) is in <(3,3)>? you know that m ≤ 9, so that's not THAT much work.

is (0,1) in <(3,3)>? is (0,2)? keep going...until you find the m that works.

1, 3

9. ## Re: Cylic

1,3? what does that mean?