Timeline for From direct sum of quotient group of a group to direct sum of the group
Current License: CC BY-SA 4.0
17 events
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| Apr 20, 2022 at 7:40 | comment | added | user110901 | @GerryMyerson, I am sorry for my description. The problem is the factorization theory of abelian groups, which deals with decomposing an abelian group into a direct sum of its subsets. If you are interested, you can read the book, Factoring Groups into Subsets, by Sándor Szabó and Arthur D. Sands. | |
| Apr 20, 2022 at 5:52 | comment | added | Gerry Myerson | @Derek, internal or external, the direct sum is a sum of subgroups, but in the example OP gives, neither $A$ nor $B$ is a subgroup. | |
| Apr 19, 2022 at 14:47 | comment | added | user110901 | @DerekHolt, yes, you are right. So, what is conditions do we need for the result to hold? For example, $H\neq Z_n$ and $H\cap A\neq \emptyset$, $H\cap B\neq \emptyset$, $((H\cap A)-(H\cap A))\cap((H\cap B)-(H\cap B))\neq \{0\}$, or others. Tanks for your answer. | |
| Apr 19, 2022 at 14:40 | comment | added | user110901 | @GerryMyerson I am sorry for my description, my means internal direct sum of two set. | |
| Apr 19, 2022 at 14:10 | comment | added | Derek Holt | @GerryMyerson Formally, there are two types of direct sum(or product), which are often used interchangeably in practice. The external direct sum of $A$ and $B$ consists of ordered pairs $(a,b)$ with co-ordinatewise addition. We say that a group $G$ is the internal direct product of $A$ and $B$ if $A$ and $B$ are subgroups of $G$ with $A+B = G$ and $A \cap B = \{0\}$. The OP seems to be talking about internal direct products sums. The external direct sum is the internal sum of the subgroups $\{(a,0):a \in A\}$ and $\{(0,b):b \in B\}$, which are often identified with $A$ and $B$. | |
| Apr 19, 2022 at 12:53 | comment | added | Gerry Myerson | OK, I guess what threw me off was I was interpreting $A\oplus B$ as $\{\,(a,b):a\in A,b\in B\,\}$, whereas what you mean is $\{\,a+b:a\in A,b\in B\,\}$. | |
| Apr 19, 2022 at 10:23 | comment | added | Derek Holt | In your question we could have $H = Z_n$, and $A$ and $B$ could be arbitrary subgroups, so the answer is no. | |
| Apr 19, 2022 at 9:59 | comment | added | user110901 | @GerryMyerson I am sorry for my description. Consider the following the example. Let $G=Z_{270}$ and $H=\{0,45,90,135,180,225\}$. Suppose $B=\{0,1,44,45,46,224,225,226,269\}$. Then $(B+H)/H=\{H, 1+H, 44+H\}$, where $1+H=\{1+h|h\in H\}$. Taking [(A+H)/H=\{H, 3+H, 6+H,9+H,12+H,15+H,18+H,21+H,24+H,27+H,30+H,33+H,36+H,39+H,42+H\},] we have $Z/H=(A+H)/H\oplus (B+H)/H$ and $Z_{270}=A\oplus B$, where $A=\{0, 3, 6,9,12,15,18,21,24,27,30,33,36,39,42\}+\{0,135\}$. For the general case, can we get the same result? From $Z_n/H=(A+H)/H\oplus (B+H)/H$ to get $Z_n=A\oplus B$. Thanks for your suggestion. | |
| Apr 19, 2022 at 9:53 | comment | added | user110901 | @GerryMyerson I am sorry for my description. Consider the following the example. Let $G=Z_{270}$ and $H=\{0,45,90,135,180,225\}$. Suppose $B=\{0,1,44,45,46,224,225,226,269\}$. Then $(B+H)/H=\{H, 1+H, 44+H\}$, where $1+H=\{1+h|h\in H\}$. Taking $(A+H)/H=\{H, 3+H, 6+H,9+H,12+H,15+H,18+H,21+H,24+H,27+H,30+H,33+H,36+H,39+H,42+H\}$, we have $Z/H=(A+H)/H\oplus (B+H)/H$ and $Z_{270}=A\oplus B$, where $A=\{0, 3, 6,9,12,15,18,21,24,27,30,33,36,39,42\}+\{0,135\}$. For the general case, can we get the same result? From $Z_n/H=(A+H)/H\oplus (B+H)/H$ to get $Z_n=A\oplus B$. Thanks for your suggestion. | |
| Apr 19, 2022 at 9:05 | comment | added | user110901 | @GerryMyerson Let $G=Z_{270}$ and $H={0,1,44,45,,46,224,225,226,269}$. | |
| Apr 19, 2022 at 8:49 | comment | added | Gerry Myerson | So, what does $A+H$ mean? What does $H\cap A$ mean? $G$ is a set of ordered pairs, so is $H$, but $A$ isn't. I still don't know what you are talking about. Maybe you have an example? | |
| Apr 19, 2022 at 7:46 | comment | added | user110901 | @GerryMyerson, I am sorry, I reworte the condition, where $H$ is a subgroup of $G$, and $H\cap A$ is a subgroup of $G$. But $A$ and $B$ are not subgroups of $G$. Thanks for your suggestions. | |
| Apr 19, 2022 at 7:41 | history | edited | user110901 | CC BY-SA 4.0 |
added 54 characters in body
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| Apr 19, 2022 at 7:05 | comment | added | Gerry Myerson | If $H$ is a subgroup of $A\oplus B$, then it's not a subset, let alone a subgroup, of $A$, nor of $B$, so it's not clear what you are talking about. | |
| Apr 18, 2022 at 17:00 | review | Close votes | |||
| May 7, 2022 at 3:04 | |||||
| S Apr 18, 2022 at 15:57 | review | First questions | |||
| Apr 18, 2022 at 19:34 | |||||
| S Apr 18, 2022 at 15:57 | history | asked | user110901 | CC BY-SA 4.0 |