Subgroup of the direct product of the rings or groups

Question

Hi I'm just consider the difference between groups and rings when it comes to direct product. And want to check this is right or not.

Let $A_i \le B_i$ [The $A_i$ is a subobject(Subring or Subgroup) of the $B$]

It is obvious that

$A_i \le B_i \Rightarrow \Pi _{i=1} ^{n} A_i \le B(=\Pi _{i=1} ^{n} B_i)$

Then question is

First)

Is it true that $A_i \le B_i \Leftarrow \Pi _{i=1} ^{n} A_i \le B(=\Pi _{i=1} ^{n} B_i)$ ?

It looks like a not true but I couldn't find any counter examples.

Second )

$\forall$ suboject of ( $A_1 \times A_2\times...\times A_n$) = (subobject of $A_1$) $\times$ (subobject of $A_2$) $\times$ .... $\times$ (subobject of $A_n$)?

i.e. Can the all sub-objects(subrings or subgroups) of the $B$ be expressed as a $\Pi _{i=1} ^{n} A_i$ ?

Third)

If not, What conditions we need that the above things are true respectively with the case of the Ring and Group?

Answer 1

The answer to the first question is also no. Let $B_1=S_4$ and $B_2=S_3$. Let $A_1=C_6$, the cyclic group of order $6$, and let $A_2=C_2$, the cyclic group of order $2$. Then $A_1\times A_2$ embeds into $B_1\times B_2$ by taking the generators $\Bigl((1,2),(1,2,3)\Bigr)$ for the cyclic group of order $6$, and the generator $\Bigl((3,4),e\Bigr)$ for the cyclic group of order $2$. This generates a subgroup that is isomorphic to $A_1\times A_2$, so identify this product with that subgroup.

However, there is no way to embed $A_1$ into either $B_1$ or $B_2$, because neither $S_4$ nor $S_3$ have elements of order $6$.

Answer 2

Contrary to what you believe, I believe that the answer to your first question is YES. Here is why. If $\pi_k: \prod_{i=1}^nB_i \to B_k$ is the projection homomorphism of the $k^{th}$ component and $\prod_{i=1}^nA_i$ is a subgroup of the direct product, then $A_k = \pi_k(\prod_{i=1}^nA_i)$ is a subgroup of $B_k = \pi_k(\prod_{i=1}^nB_i)$ (since $\pi_k$ is a homomorphism). This also works for rings.

The answer to the second question is no. Consider the diagonal subgroup. Namely, if we consider the group $\mathbb{Z} \times \mathbb{Z}$, then $\{(a,a):a\in \mathbb{Z}\}$ is a subgroup of $\mathbb{Z} \times \mathbb{Z}$, but is not in the form you described. This also works for rings.

As regards to your third question, I am not aware of any conditions (while I am sure there are some). For a very specific case, I think that you can prove that if you have a finite collection of groups whose orders are pairwise relatively prime, then any subgroup of the direct product will be a direct product of subgroups for each group separately. However, like I said this is a very specific case and does not completely answer the third question.