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Can we find all group homomorphism from $\mathbb{Z}^n$ to $\mathbb{Z}^n$?

For such a map surjective always imply isomorphism (like $\mathbb{Z}$)?

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  • $\begingroup$ do you mean the free abelian product? $\endgroup$ – Daniel Valenzuela Aug 7 '14 at 10:53
  • $\begingroup$ usual direct sum of n copy of integers $\endgroup$ – jeevan Aug 7 '14 at 10:54
  • $\begingroup$ Look at the proof of how linear maps of vector spaces (modules over a field) are in one-to-one correspondence to matrices (with fixed basis). The same will work here. $\endgroup$ – Daniel Valenzuela Aug 7 '14 at 10:56
  • $\begingroup$ what answer should I expect? $\endgroup$ – jeevan Aug 7 '14 at 11:08
  • $\begingroup$ you know how all linear maps look like $\endgroup$ – Daniel Valenzuela Aug 7 '14 at 11:10
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if the epic morphism were not monic then it would have a non-trivial kernel $K$. then, we would have $Z^n \cong Z^n \oplus K$

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  • $\begingroup$ can you explain in little detail? $\endgroup$ – jeevan Aug 7 '14 at 11:04
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    $\begingroup$ you need to tell us what you know. The statement is clear when you know about split exact sequences. Otherwise you may know about the isomorphism theorem, so that a surjective map $Z^n \to Z^n$ induces an isomorphism $Z^n/K \to Z^n$ where $K$ is the kernel. Since the rank is an invariant (preserved by isomorhisms) as soon as the kernel $K$ is non trivial, the map would not be an isomorphism. Therefore surjectiveness implies trivial kernel; hence an epimorhism is automatically an isomorphism. $\endgroup$ – Daniel Valenzuela Aug 7 '14 at 11:08
  • $\begingroup$ thx Dan. i am not on very sure ground as yet! my thought was that $Z^n$ is a projective $Z$-module and that would ensure that an exact sequence splits. however i am not very clear on what "projective" means. from the little i have read it seems to be to do with having a basis. your comment suggests it is the finite rank which is the crux of the matter. could you suggest a good short ref which might help me to sharpen up my understanding? $\endgroup$ – David Holden Aug 7 '14 at 11:16
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$\operatorname{Hom}_{\mathbb Z}(\mathbb Z^n,\mathbb Z^n)\simeq\operatorname{Hom}_{\mathbb Z}(\mathbb Z,\mathbb Z^n)^n\simeq\mathbb Z^{n^2}$. In fact, a group homomorphism from $\mathbb Z^n$ to $\mathbb Z^n$ corresponds to a matrix in $M_{n}(\mathbb Z)$.

Every surjective endomorphism of $\mathbb Z^n$ is an isomorphism: if $A\in M_n(\mathbb Z)$ is the corresponding matrix of a surjective endomorphism of $\mathbb Z^n$, then there exists $B\in M_n(\mathbb Z)$ such that $AB=I_n$. Now, by taking the determinant, it follows that $A$ is invertible. (For a more general case see here.)

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  • $\begingroup$ thx @user26857, that link is a great introduction to Nakayama's Lemma $\endgroup$ – David Holden Aug 17 '14 at 8:48

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