- #1
ehrenfest
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[SOLVED] finitely generated abelian groups
My book states that
[tex] (\mathbb{Z} \times\mathbb{Z} \times\cdots \times \mathbb{Z})/(d_1\mathbb{Z} \times d_2\mathbb{Z} \times \cdots d_s\mathbb{Z} \times {0} \times \cdots \times {0}) [/tex]
is isomorphic to
[tex] \mathbb{Z}_{d_1} \times\cdots \times \mathbb{Z}_{d_s} \times \mathbb{Z} \times\cdots \times \mathbb{Z}[/tex]
with absolutely no explanation of any sort. I don't know why this is so obvious to everyone because it is NOT TRUE that you can just mod out things by there components as you can clearly see from the fact that: (Z_4 \times Z_6)/<(2,3)> is not isomorphic to Z_2 \times Z_3.
I have absolutely no idea what thought process went into that statement above and why they think they can just mod things out by there components when that is just wrong.
Homework Statement
My book states that
[tex] (\mathbb{Z} \times\mathbb{Z} \times\cdots \times \mathbb{Z})/(d_1\mathbb{Z} \times d_2\mathbb{Z} \times \cdots d_s\mathbb{Z} \times {0} \times \cdots \times {0}) [/tex]
is isomorphic to
[tex] \mathbb{Z}_{d_1} \times\cdots \times \mathbb{Z}_{d_s} \times \mathbb{Z} \times\cdots \times \mathbb{Z}[/tex]
with absolutely no explanation of any sort. I don't know why this is so obvious to everyone because it is NOT TRUE that you can just mod out things by there components as you can clearly see from the fact that: (Z_4 \times Z_6)/<(2,3)> is not isomorphic to Z_2 \times Z_3.
I have absolutely no idea what thought process went into that statement above and why they think they can just mod things out by there components when that is just wrong.
Homework Equations
The Attempt at a Solution
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