
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.
Could anyone help me? please!!

Re: Cylic
1. this is just straightforward 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?

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?

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)>?

Re: Cylic

Re: Cylic

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.

Re: Cylic

Re: Cylic
1,3? what does that mean?