We know that for a cyclic group $G$, if $G=A\oplus B$, then for some subgroups $H$ of $G$, We have $G/H=(A+H)/H\oplus (B+H)/H.$ But, if we know that for a subgroup $H$ of $G$, $G/H=(A+H)/H\oplus (B+H)/H$, where $H\cap A$ is a subgroup of $G$. Then what conditions do we need to make $G=A\oplus B$?
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$\begingroup$ 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. $\endgroup$– Gerry MyersonCommented Apr 19, 2022 at 7:05
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$\begingroup$ @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. $\endgroup$– user110901Commented Apr 19, 2022 at 7:46
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$\begingroup$ 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? $\endgroup$– Gerry MyersonCommented Apr 19, 2022 at 8:49
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$\begingroup$ @GerryMyerson Let $G=Z_{270}$ and $H={0,1,44,45,,46,224,225,226,269}$. $\endgroup$– user110901Commented Apr 19, 2022 at 9:05
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$\begingroup$ @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. $\endgroup$– user110901Commented Apr 19, 2022 at 9:53
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