- #1

markiv

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In other words, we know that [tex]R/I[/tex] is isomorphic to some ideal in [tex]R[/tex], call this [tex]J[/tex]. It's clear that [tex]I \cap J = 0[/tex], but does [tex]I + J = R[/tex] ?

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- Thread starter markiv
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- #1

markiv

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In other words, we know that [tex]R/I[/tex] is isomorphic to some ideal in [tex]R[/tex], call this [tex]J[/tex]. It's clear that [tex]I \cap J = 0[/tex], but does [tex]I + J = R[/tex] ?

- #2

Hurkyl

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What is true? It is certainly not true that [itex]G \cong G/H \times H[/itex] for all abelian groups G and subgroups H -- e.g. let G be the cyclic group of 4 elements....I know it's true for finitely-generated Abelian groups

- #3

markiv

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Thanks!

- #4

markiv

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What is true? It is certainly not true that [itex]G \cong G/H \times H[/itex] for all abelian groups G and subgroups H -- e.g. let G be the cyclic group of 4 elements....

I'm just trying to understand why I thought this, but if [tex]H[/tex] is Sylow in [tex]G[/tex], then [tex]G \cong G/H \oplus H[/tex], right? Still assuming [tex]G[/tex] is FG Abelian, of course.

- #5

Hurkyl

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I never really remember the Sylow stuff. I have three other guesses as to why you might have thought it, though:I'm just trying to understand why I thought this

- It's true for vector spaces.
- It's "often" true for abelian groups (e.g. if the quotient is torsion free)
- There is an isomorphism of
*sets*between G and H (+) G/H

(For the third point, I'm pretty sure I'm using the axiom of choice)

Or, maybe you're partially remembering a general theorem:

If there is a group homomorphism f : G/H --> G with the property that f(x) = x (mod H), then G is isomorphic to H (+) G/H

- #6

markiv

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- #7

markiv

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Wait actually... FG Abelian groups are nilpotent, and a group is nilpotent if and only if it is a direct sums of its Sylow subgroups. So [tex]G \cong G/H \oplus H[/tex] if [tex]H[/tex] is a Sylow subgroup of nilpotent group [tex]G[/tex] ?

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