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Suppose we have a homomorphism defined as $\phi: \mathbb{Z}\to \mathbb{Z_3}$, where $\mathbb{Z_3}$ is a group under addition. Then $\ker{\phi}$ will be the set $3\mathbb{Z}=\{0, 3, 6, ..., 27, ...\}$. Now let's take some element in the kernel of $\phi$ (say, $3$) and take the product with the kernel of $\phi$. We will then have $3+\ker{\phi} = \{3, 6, 9, ...,\}$. Basically, this operation appears to eliminate all elements of the kernel, which precede $3$.

I'm wondering what I'm not seeing here, to be able to be satisfied that all cosets of the kernel of some homomorphism must be the kernel of the homomoprhism?

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Note that $3\mathbb{Z}=\{...,-9,-6,-3,0,3,6,9,...\}$, so $3+\ker(\phi)=\{...,-6,-3,0,3,6,...\}$.

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$3\mathbb Z=\{\dots, -9,-6,-3,0,3,6,9,\dots\}$. The integers wouldn't be a group under addition if they didn't have negative numbers.

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