Exact sequence with finite groups: $0 \to A \xrightarrow[]{\alpha} \mathbb{Z}^d \xrightarrow[]{\beta} \mathbb{Z} \xrightarrow[]{\gamma} B \to0$.

Question

I have this exact sequence of abelian groups: $$0 \to A \xrightarrow[]{\alpha} \mathbb{Z}^d \xrightarrow[]{\beta} \mathbb{Z} \xrightarrow[]{\gamma} B \to0$$ with $A$ and $B$ finite abelian groups and $d \in \mathbb{N}$.

I would like to conclude $d=1$, but I have no idea how could use the information $A,B$ are finite. The only thing I noticed that $\gamma$ cannot be injective, but I don't know how to proceed. I'm a bit rusty in Algebra...

Could someone give some help?

Thanks!

Answer 1

You can first conclude that $\alpha = 0$: if not, $A \neq 0$, and any non-zero element of $A$ would be sent by $\alpha$ to a torsion element of $\mathbb{Z}^d$, but no such exist. So $\alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $\beta: \mathbb{Z}^d \to \mathbb{Z}$.

But if $d\geq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $\beta(\beta(e_1)e_2) = \beta(e_1)\beta(e_2) = \beta(\beta(e_2)e_1)$, so $\beta$ cannot be an injection, so indeed, $d \leq 1$ (and $d \neq 0$ else $\gamma$ would be an isomorphism).

Answer 2

Since $A$ is finite then there is no nontrivial homomorphism $A\to\mathbb{Z}^d$. Meaning $\alpha=0$. This implies that $\ker\beta=0$ and so $\beta$ is injective. The case when $\beta$ is injective is only possible when $d\leq 1$.

The case when $d=0$ is impossible because that would imply that $\gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.

EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $\alpha=0$ but $0 \to A \xrightarrow[]{\alpha} \mathbb{Z}^d$ implies it is injective.

Answer 3

Tensoring with the flat $\mathbb{Z}$-module $\mathbb{Q}$ gives an exact sequence $0 \to \mathbb{Q}^d \to \mathbb{Q} \to 0$.