Is the Direct Product of Two Finite Cyclic Groups Cyclic

[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.