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What does Z6 x Z3 look like?by scharl4
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#1
Jun1812, 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!! 


#2
Jun1812, 09:44 AM

P: 428




#3
Jun1812, 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]? 


#4
Jun1812, 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.



#5
Jun1912, 12:33 AM

P: 428

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. 


#6
Jun1912, 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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