Question

how we know that factor group (Z(mod 4)xZ(mod 6))/<(0,2)> is isomorphic to Z(mod 4)xZ(mod 2)

Answer 1

Alright, the factor group is just the group of all cosets, right? So, first, let's look at what our subgroup <(0,2)> looks like. It's going to have three elements: (0,0), (0,2), (0,4). Right away, it should be really clear that the first half of the direct product in the solution will remain Z/4. Every element in our subgroup has a 0 first term, so we'll have four different partitions with respect to that term. The second part isn't much harder, you can see that the cosets will either have 1,3,5 or 0,2,4.

Here's a way to think about it that might make it easier. Break the direct product up into its two halves. Then you just have (Z/4)/{0} (which is obviously and trivially Z/4), and (Z/6)/<2>. Quick way on this is to note that the subgroup generated by 2 has order 3, Z/6 has order 6, so the quotient group must have order 2, so you're clearly going to have Z/2.

Does that make sense?

Answer 2

If you still need to see more, this group actually wouldn't be too terribly hard to write out. A bit tedious, but doable.

Answer 3

@experimentX ,do you have any other method?i still don't understand the answer given...

Answer 4

I will try to give a better explanation after I get back from work, about 8 hours from now.

Answer 5

ok..thanks..

Answer 6

i think it is easier if we list first all possible abelian groups of order 8..then we decide which group isomorphic to this factor group...

Answer 7

Okay. Let's look at it very explicitly. We have the group \(\mathbb Z /4\times\mathbb Z /6.\) So the elements of that group are of the form \((0\dots3,0\dots5).\) We are taking the quotient group, using the subgroup generated by \((0,2).\) It should be pretty clear that that subgroup has three elements: \((0,0), (0,2), (0,4).\) Hopefully that is clear, if not, let me know. Anyway, moving on from that, when we look at the quotient group, we are simply looking at all of the cosets of that subgroup. So, if we call that subgroup \(H\), we are looking at the group of \(xH\) where \(x\in G\), with \(G\) being our original group \(\mathbb Z /4\times\mathbb Z /6.\) Let's look at some examples of cosets, and hopefully that will make this more clear.

The trivial coset is \(eH\), where \(e=(0,0).\) This will of course just give us back the subgroup \(H,\ \{(0,0), (0,2),(0,4)\}.\) Next, let's look at the coset for the element \((0,1).\) We multiply this by each element in \(H\) and get \(\{(0,1),(0,3),(0,5)\}.\) Now, let's see the coset for the element \((0,2).\) We multiply by each element in \(H\) and get \(\{(0,2),(0,4),(0,0)\}.\) This is the same as \(\{(0,0),(0,2),(0,4)\}.\) So, the coset where \(x=(0,2)\) is the same as the coset where \(x=(0,0).\) Next, let's look at the coset where \(x=(0,3).\) We end up with \(\{(0,3),(0,5),(0,1)\}.\) But that is the same as \(x=(0,1).\) So you see, when the first element of \(x\) is the same, we have basically two options. Either we end up with \((x,0),(x,2),(x,4)\), or we have \((x,1),(x,3),(x,5).\)

For the first element, we can have any of the four elements of our original group. 0, 1, 2, or 3. So, we end up with \(\mathbb Z/4\times\mathbb Z/2.\) Does this make it clearer at all? Let me know which specific parts you are having trouble with.