I Enumerating the cosets of a kernel

Mr Davis 97

Suppose that we have, for purposes of example, the homomorphism $\pi : \mathbb{R}^2 \to \mathbb{R}$ such that $\pi((x,y)) = x+y$. We see that $\ker(\pi) = \{(x,y)\in \mathbb{R}^2 \mid x+y=0\}$. How can we enumerate all of the cosets of the kernel? My thought was that of course as we range $g$ over $G$ we look at $g\ker (\pi)$ we get all of the cosets, but how can I find a subset of $I \subseteq G$ such each element of $I$ gives a new coset when multiplied by the kernel?

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fresh_42

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Suppose that we have, for purposes of example, the homomorphism $\pi : \mathbb{R}^2 \to \mathbb{R}$ such that $\pi((x,y)) = x+y$. We see that $\ker(\pi) = \{(x,y)\in \mathbb{R}^2 \mid x+y=0\}$. How can we enumerate all of the cosets of the kernel? My thought was that of course as we range $g$ over $G$ we look at $g\ker (\pi)$ we get all of the cosets, but how can I find a subset of $I \subseteq G$ such each element of $I$ gives a new coset when multiplied by the kernel?
I see only additive groups here, so it has to be $g+\operatorname{ker}(\pi)$. Moreover we have an epimorphism, so we get an induced isomorphism $\bar{\pi}\, : \,\mathbb{R} \cong \mathbb{R}^2/\operatorname{ker}(\pi)$ which means, all elements are of the form $g+\mathbb{R}$ where $\mathbb{R}=\operatorname{ker}(\pi)$ is the diagonal, which is shifted upwards ($g>0$) or downwards ($g<0$) to cover the entire plane by copies of said diagonal.

"Enumerating the cosets of a kernel"

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