- #1
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If one has a short exact sequence 0-->A-->B-->C-->0 of finitely generated abelian groups, how does one show that rank(B)=rank(A)+rank(C) ?
We have that A embeds in B and C is isomorphic to B/A. The natural thing to try to use I think is the uniqueness of the decomposition of a finitely generated abelian group into cyclic groups, but how?
Like, if [tex]\phi :B\cong \mathbb{Z}^r\oplus\mathbb{Z}_{n_1}\oplus\ldots\mathbb{Z}_{n_b}[/tex], even though [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] for some isomorphism, there is no guarantee that the embedded A (in B) will map to [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] under [itex]\phi[/itex]. Or is there?
We have that A embeds in B and C is isomorphic to B/A. The natural thing to try to use I think is the uniqueness of the decomposition of a finitely generated abelian group into cyclic groups, but how?
Like, if [tex]\phi :B\cong \mathbb{Z}^r\oplus\mathbb{Z}_{n_1}\oplus\ldots\mathbb{Z}_{n_b}[/tex], even though [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] for some isomorphism, there is no guarantee that the embedded A (in B) will map to [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] under [itex]\phi[/itex]. Or is there?