What does Z6 x Z3 look like?


by scharl4
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scharl4
scharl4 is offline
#1
Jun18-12, 09:04 AM
P: 6
1. The problem statement, all variables and given/known data
Find all cyclic subgroups of Z6 x Z3.

2. Relevant equations



3. The attempt at a solution
I understand how to find a cyclic subgroup of a simpler group such as Z4, but having trouble understanding what subgroups look like in a direct product of integer spaces, let alone cyclic subgroups. Also, having trouble understanding what makes a direct product subgroup cyclic. Is it cyclic when (a1, a2)^n = (e1, e2)? Please help!!
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algebrat
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#2
Jun18-12, 09:44 AM
P: 428
Quote Quote by scharl4 View Post
1. The problem statement, all variables and given/known data
Find all cyclic subgroups of Z6 x Z3...

...Also, having trouble understanding what makes a direct product subgroup cyclic. Is it cyclic when (a1, a2)^n = (e1, e2)? Please help!!
A group is cyclic if it is generated by a single element. So find subgroup generated by each element (some elements might generate the same subgroup).
tonit
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#3
Jun18-12, 11:17 AM
P: 55
Direct product of two groups [itex]G[/itex] and [itex]H[/itex], is the group [itex]G\times H = \{ (g,h) | g \in G, h \in H \}[/itex].

If [itex]*[/itex] is the operation of G and H, [itex](g,h)*(g_1,h_1) = (g*g_1,h*h_1)[/itex]. Similarly the inverse [itex](g,h)^{-1} = (g^{-1},h^{-1})[/itex].

Now can you find any element [itex](g,h) \in \mathbb{Z}_6\times \mathbb{Z}_3[/itex] such that each element in [itex]\mathbb{Z}_6\times \mathbb{Z}_3[/itex] can be represented in the form [itex](g^n,h^n)[/itex]?

scharl4
scharl4 is offline
#4
Jun18-12, 08:00 PM
P: 6

What does Z6 x Z3 look like?


Ok so I think I found all the subgroups generated by each element of Z6 x Z3. It looks like (0,0), (0,1), (1,0), (1,1), (3,0), (3,1), (4,0), (4,1), (5,0), (5,1) each generate only themselves, while each of the other elements of Z6 x Z3 only generate 2 elements. So, it appears that there are no cyclic subgroups, unless i computed the subgroups incorrectly.
algebrat
algebrat is offline
#5
Jun19-12, 12:33 AM
P: 428
Quote Quote by scharl4 View Post
Ok so I think I found all the subgroups generated by each element of Z6 x Z3. It looks like (0,0), (0,1), (1,0), (1,1), (3,0), (3,1), (4,0), (4,1), (5,0), (5,1) each generate only themselves, while each of the other elements of Z6 x Z3 only generate 2 elements. So, it appears that there are no cyclic subgroups, unless i computed the subgroups incorrectly.
If your notation includes zero, you must be using the additive notation. When you say (0,1) only generates itself, do you mean it generates a subgroup of one element? There's two things wrong with that.

One is, every group, and subgroup, must contain the identity by definition. Under your implicit choice of addition as the operation, the identity is (0,0).

The second problem is, 5 is relatively prime to 6, so for instance 5+...+5=5*5=25=1+24=1 mod 6. So (5,0) generates the same group (1,0) does.

Perhaps you do not know what it means for an element to generate a subgroup. Under addition, all multiples, under multiplication, all powers (a sometimes tricky distinction).

Please be more careful, and propose different groups.
scharl4
scharl4 is offline
#6
Jun19-12, 11:08 AM
P: 6
OK thanks. I got a little confused because the other post said I should generate by using (a^n, b^n). I think I have my subgroups correct now.


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