Question

how to find the group that isomorphic to this factor group, (Z4xZ6)/<(0,2)>

Answer 1

I'm sorry, my abstract algebra is a bit too rusty since I'm running in circles here. You should try http://math.stackexchange.com/ .

This site is good, but doesn't have many college students or professors.

Answer 2

I mean, http://math.stackexchange.com/ has more subject experts than here. OpenStudy is really just a study hall.

Answer 3

ok..thank you..

Answer 4

Hmm...well you know that:\[\mathbb{Z}_4\times \mathbb{Z}_6\]has 24 elements in it, and the cyclic subgroup generated by (0,2) has only 2 elements:\[\langle(0,2)\rangle=(0,2),(0,4)\]So the factor group will have order 24/2 = 12. Since the original group is abelian, the factor group will also be abelian, so we can use the Fundamental Theorem of Abelian Groups to say this group must be isomorphic to either:\[\mathbb{Z}_{12},\mathbb{Z}_3\times \mathbb{Z}_4,\mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_2\]All that remains is to find out wihch of these three groups it is. Might have to do some direct calculation of the factor group. Or possibly eliminate some answers.

Answer 5

<(0,2)>={(0,0),(0,2),(0,4)}...am i right?so,the order of factor group is 8..then?

Answer 6

ah yes. my bad. i have that on my white board but i didnt put it up here >.<

Answer 7

So then its 24/3 = 8, and your choices are:\[\mathbb{Z}_8,\mathbb{Z}_4\times\mathbb{Z}_2,\mathbb{Z}_2\times,\mathbb{Z}_2\times\mathbb{Z}_2\]

Answer 8

er, that last one shouldnt have the comma

Answer 9

\[\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\]

Answer 10

the answer given is:
the first factor Z4 of Z4 x Z6 is left alone.The Z6 factor,on the other hand,is essentially collapsed by a subgroup of order 3,giving a factor group in the second factor of order 2 that must be isomorphic to Z2.Thus the factor group is isomorphic to Z4 x Z2.
can you explain this answer...

Answer 11

I cant really explain that =/ thats not how I would have done it. I would have started calculating the order of some elements. It would be easy to find one element that has order greater than 2, so that would eliminate Z2 x Z2 x Z2. Then I would show that there is no element of order 8, so the answer couldnt be Z8.