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In a course of group theory an exercise is left with no indications :

Let $A$ be an abelian group of finite type and $a_1,\dots ,a_n$ a $\mathbb{Z}$-basis of $A$. Let $f : A \to A$ be a group homomorphism. Les $M=(m_{ij})$ be the matrix (invertible) defined by $\displaystyle f(a_j)=\sum_{i=1}^n m_{ij}a_i$ with $\det (M)\neq 0$. Show that $[A:\operatorname{Im}(f)]=|\det M |$, i.e the index of $\operatorname{Im}(f)$ in $A$ as a subgroup is precisely $\det (M)$.

I am not able to prove this fact, and need a bit of help. In fact I do not see where I can make $\det (M)$ appear. Can you give me some hints ?

Thank you

user26857
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J.Mayol
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1 Answers1

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The idea is that first you prove it in case of diagonal $M$, and then you construct a homomorphism $h: A \to A$ such that $\operatorname{im} f = \operatorname{im} h$, and $h$ is given by a diagonal matrix that is similar to $f$, so $\det f = \det h$.

The construction is basically the same as in the proof of the classification of finite abelian groups (or modules over a PID).

xyzzyz
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    You won't exactly have $\operatorname{im} f = \operatorname{im} h$, but they would be related by an automorphism. – arkeet Oct 18 '16 at 20:59
  • Thank you for the fast answer ! The result seems way easier when $M$ is diagonal, but I didn't tought about it, thanks ! – J.Mayol Oct 18 '16 at 21:03