Direct Product of Cyclic Groups

[Bashyboy]

Hello everyone,

I was wondering if the following claim is true:

Let ##G_1## and ##G_2## be finite cyclic groups with generators ##g_1## and ##g_2##, respectively. The group formed by the direct product ##G_1 \times G_2## is cyclic and its generator is ##(g_1,g_2)##.

I am not certain that it is true. If I make the following stipulation

Let ##G_1## and ##G_2## be finite cyclic groups with generators ##g_1## and ##g_2##, respectively, and the group formed by the direct product ##G_1 \times G_2## is cyclic, then it has the generator ##(g_1,g_2)##.

this might be true. However, I would like to hear from you before I try to go prove something that is false.

 

[WWGD]

The product is cyclic iff the orders are relatively-prime. In the direct product, you have:

## |G_1 \times G_2|=|G_1| \times |G_2| ## . Then, if you can find an element in the product with that
order, you are done.

 

[Bashyboy]

More importantly, I am interested in knowing if the generator of ##G_1 \times G_2## is based off the generators of the individual groups ##G_1## and ##G_2##.

Or is this not true in general?

 

[WWGD]

Notice that the generator of a cyclic group is not necessarily unique.

RE your question on relation between individual generators and generators of the product: yes, it is. Let ## g_1, g_2 ## be generators for ##G_1, G_2 ## respectively. Then there are positive integers ##m,n## with

##g_1^n=e_{G_1} , g_2^m = e_{G_2}##. What is then the order of ## (g_1, g_2)## ?

 

[Bashyboy]

Let's see if I understand this correctly: Let ##|G_1| = x## and ##|G_2| = y##, and let both be both cyclic with the generators alluded in the above posts. If this is so, then ##\langle g_1 \rangle = G_1 \implies |\langle g \rangle | = x##, with a similar thing being true of ##G_2##. Furthermore, ##|G_1 \times G_2 | = xy##. There is a theorem that I proved which shows that the direct product of two subgroups produces a subgroup of the direct product of the larger group(s). Therefore, ##\langle g_1 \rangle \times \langle g_2 \rangle## is a subgroup, and its order is ##| \langle g_1 \rangle \times \langle g_2 \rangle | = xy##. Thus, the element ##(g_1,g_2)## has an order of ##xy##, and it must generate ##G_1 \times G_2##.

Does this sound right?

 

[TheMathSorcerer]

Let's see if I understand this correctly: Let ##|G_1| = x## and ##|G_2| = y##, and let both be both cyclic with the generators alluded in the above posts. If this is so, then ##\langle g_1 \rangle = G_1 \implies |\langle g \rangle | = x##, with a similar thing being true of ##G_2##. Furthermore, ##|G_1 \times G_2 | = xy##. There is a theorem that I proved which shows that the direct product of two subgroups produces a subgroup of the direct product of the larger group(s). Therefore, ##\langle g_1 \rangle \times \langle g_2 \rangle## is a subgroup, and its order is ##| \langle g_1 \rangle \times \langle g_2 \rangle | = xy##. Thus, the element ##(g_1,g_2)## has an order of ##xy##, and it must generate ##G_1 \times G_2##.

Does this sound right?

The order need not be ##xy##. Take ##Z_2## which is cyclic with generator ##1##. Then in the group ##Z_2 \times Z_2##, the element ##(1,1)## has order two since ##(1, 1) \neq (0,0)##, and ##(1, 1) + (1, 1) = (0, 0)##, yet ##Z_2 \times Z_2## has order ##4##, so ##(1,1)## does not generate ##Z_2 \times Z_2##. Note ##Z_2 \times Z_2## is isomorphic to the Klein 4-group which is not cyclic. I think perhaps a look at the previous post is a good idea as it really says it all. What is the order of ##(g_1, g_2)##? It is worth looking at. LaTeX fixed, thanks Greg:)

 

[Greg Bernhardt]

Apologies for the lack of LaTeX, still figuring out how to use it here.

Needs to be updated, but still good
https://www.physicsforums.com/help/latexhelp/

 

[WWGD]

No, the order of the product equals the product of the orders only if the orders are relatively-prime to each other.
Otehrwise, you use the LCM.