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I would like to ask if two different homomorphisms can share the same kernel.

For instance for the kernel $n \mathbb{Z} $, is it possible to come up with homomorphisms other than the function mapping integers to residue classes modulo $n$? Thanks.

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No, a homomorphism is not uniquely determined by its kernel. Consider the following two homomorphisms from $\mathbb{Z}_2$ to $\mathbb{Z}_2\times\mathbb{Z}_2$: one sending $1$ to $(0,1)$ and the other sending $1$ to $(1,0)$. They're both homomorphisms with the same kernel to the same group, but they are different homomorphisms.

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  • $\begingroup$ Saying "no" is a bit confusing because the question in the body is the negation of the question in the title, and it isn't clear which you are answering at first. $\endgroup$ Dec 5 '16 at 5:10
  • $\begingroup$ Thanks - I'll edit that in. $\endgroup$ Dec 5 '16 at 5:10
  • $\begingroup$ Thank you for the answer. So is it possible to propose a different homomorphic map satisfying the kernel constraints for the example in the question? $\endgroup$ Dec 5 '16 at 5:26
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    $\begingroup$ Well, we can trivially let $\phi_1,\phi_2: \mathbb{Z}\rightarrow \mathbb{Z}_n\times\mathbb{Z}_n$, like above, where $\phi_1$ maps $1$ to $(\bar{1},0)$ and $\phi_2$ maps $1$ to $(0,\bar{1})$. The thing that's important, though, is that by the first isomorphism theorem, for any $\phi$ from $\mathbb{Z}$ to another group, $\phi$ a homomorphism, Im $\phi\cong \mathbb{Z}_n$. $\endgroup$ Dec 5 '16 at 5:38
  • $\begingroup$ @AshwinTrisal Hi, what if we pose an additional condition that the homomorphism is into a field? $\endgroup$
    – Sam Wong
    Sep 21 '18 at 6:43
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Unless $G$ has a "simple" structure, there are many isomorphisms $:G \to G$ and they all have a kernel of $\{e_G\}$.

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The identity homomorphism and the homomorphism $n\mapsto -n$ from $\mathbb{Z}$ to $\mathbb{Z}$ have ...?... kernel.

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  • $\begingroup$ Yes, they do have kernel :) $\endgroup$
    – Wojowu
    Dec 5 '16 at 13:22
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    $\begingroup$ I think $n\mapsto -n$ is not a homomorphism. For example, consider $-1$ and $-3$. The product of them is $3$, and is mapped to $-3$. But the product of their images is $3$. $\endgroup$
    – Sam Wong
    Sep 21 '18 at 10:52
  • $\begingroup$ @SamWong The tags indicate that the question is about group homomorphisms. Hence we are talking about the group structure $[\mathbb Z, +]$. $\endgroup$ Jul 15 '20 at 22:30
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Think of $k\mapsto\exp(\frac{2k\pi}n\mathbf i):\Bbb Z\to\Bbb C^\times$ which has kernel $n\Bbb Z$.

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