# Enumerating the cosets of a kernel

• I
• Mr Davis 97
In summary, the conversation discusses a homomorphism and its kernel, and how to enumerate all of the cosets of the kernel. The speaker suggests using an epimorphism to find a subset of elements that can generate all cosets when multiplied by the kernel. Additionally, the speaker notes that the groups involved are only additive.

#### 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?

Mr Davis 97 said:
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.