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


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 $\Pi _{i=1} ^{n} A_i$ ?


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

  • $\begingroup$ Suggestion: please remove the boldface type. It's bad for the same reason that writing in ALL CAPS is bad, it's like you are shouting at us. $\endgroup$ – Lee Mosher Jul 3 at 22:47
  • $\begingroup$ Thanks, Mr. Mosher. I just edited. $\endgroup$ – se-hyuck yang Jul 4 at 0:45
  • $\begingroup$ At the very least, you need to allow for a re-ordering of the $B_i$ before you include the $A_i$ into them. $\endgroup$ – Arturo Magidin Jul 4 at 1:08

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 $6$.

  • $\begingroup$ Your subgroup $A_1$ is not isomorphic to $C_6$, rather $D_6$ (or $S_3$ if you like). $\endgroup$ – Mike Jul 4 at 2:50
  • $\begingroup$ @Arturo Magidin, Like the Mike said It looks like a $\Bigl((1,2),(1,2,3)\Bigr)$ generates the $S_3$ not $C_6$. Could you please more explain why does generates the $C_6$? So, does we need the other generator for $C_6$ not $\Bigl((1,2),(1,2,3)\Bigr)$? Then Your counterexample definetly correct, I thought. $\endgroup$ – se-hyuck yang Jul 4 at 5:05
  • 1
    $\begingroup$ @Mike: You’re incorrect. Both $A_i$ are cyclic. Again: we have $S_4\times S_3$. The elements are ordered pairs, with the first entry in $S_4$ and the second entry in $S_3$. My single element is an ordered pair, $(\sigma,\tau)$, where $\sigma$ is the permutation $(1,2)$ in $S_4$, and $\tau$ is the permutation $(1,2,3)$ in $S_4$. It cannot be $S_3$ because it’s a single element. The subgroup consists of the elements $(e,e)$, $(\sigma,\tau)$, $(e,\tau^2)$, $(\sigma,e)$, $(e,\tau)$, and $(\sigma,\tau^2)$, which are the powers of $(\sigma,\tau)$. $\endgroup$ – Arturo Magidin Jul 4 at 6:56
  • 1
    $\begingroup$ @se-hyuckyang: No, Mike is wrong and so are you. It’s not $\langle (1,2),(1,2,3)\rangle$ (the subgroup generated by $(1,2)$ and $(1,2,3)$). It’s the cyclic subgroup generated by a single element of $S_4\times S_3$ of the form $(\sigma,\tau)$, $\sigma=(1,2)\in S_4$, $\tau=(1,2,3)\in S_3$. $\endgroup$ – Arturo Magidin Jul 4 at 6:58

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.

  • 1
    $\begingroup$ You seem to be assuming that the inclusion is “nice” relative to the projections. Why can you do that? That is, you are assuming that the embedding of the product of the $A_i$ into that of the $B_j$ will be such that the $j$th projection of the latter extends that of the former. I don’t think you can just assume that. $\endgroup$ – Arturo Magidin Jul 4 at 1:03
  • $\begingroup$ @ArturoMagidin I do not understand. The image of a subgroup is a subgroup. Isn't my method valid by that theorem? $\endgroup$ – Mike Jul 4 at 1:14
  • 1
    $\begingroup$ Here’s how to clarify: you have two different products. Let $p_j$ be the projection onto the $A_j$ from $\prod A_i$; let $\pi_j$ be the projection onto $B_j$ from $\prod B_j$. You have an inclusion $\prod A_i\subseteq \prod B_j$, but this is just an embedding. You are asuming that $\pi_k|_{\prod A_i} = p_k$. But that is by no means a given. For a trivial example suppose $B_1=C_2$, $B_2=C_4$, $A_1=C_4$, $A_2=\{e\}$. The embedding does not satisfy the assumption you have. Worse, you could embedd the product of $A_1\times A_2$ as the subgroup generated by $(1,1)$. Your argument would not work $\endgroup$ – Arturo Magidin Jul 4 at 1:18
  • $\begingroup$ @ Arturo Magidin What is $C_2,C_4$? $\endgroup$ – Mike Jul 4 at 1:21
  • 1
    $\begingroup$ @ArturoMagidin I just do not understand. I will have to chat about this with somebody. Looking at your profile, you have been doing this much longer than me, so I am assuming you are correct (in which case I have to figure out what is wrong with my thinking). Thank you for bringing this to my attention. $\endgroup$ – Mike Jul 4 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.