Skip to main content
MathOverflow

Return to Question

added 54 characters in body
Source Link
user110901
user110901
  • 21
  • 2

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\oplus B/H.$$G/H=(A+H)/H\oplus (B+H)/H.$ But, if we know that for a subgroup $H$ of $G$, $G/H=A/H\oplus B/H.$$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 $G=A\oplus B$?

We know that for a cyclic group $G$, if $G=A\oplus B$, then for some subgroups $H$ of $G$. We have $G/H=A/H\oplus B/H.$ But, if we know that for a subgroup $H$ of $G$, $G/H=A/H\oplus B/H.$ Then what conditions do we need to make $G=A\oplus B$?

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 $G=A\oplus B$?

Source Link
user110901
user110901
  • 21
  • 2

From direct sum of quotient group of a group to direct sum of the group

We know that for a cyclic group $G$, if $G=A\oplus B$, then for some subgroups $H$ of $G$. We have $G/H=A/H\oplus B/H.$ But, if we know that for a subgroup $H$ of $G$, $G/H=A/H\oplus B/H.$ Then what conditions do we need to make $G=A\oplus B$?