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I have a question refer to The first theorem of the isomorphisms. The theorem states that if we have a $$\dot G \cong \frac{G}{\ker(\varphi)}$$ where the co-domain $\dot G$ will be isomorphic to the quotient group $G$ by the $\ker(\varphi)$ and $\frac{G}{\ker(\varphi)}$ from my understanding contains all equivalence classes (cosets) generated by the subgroup $\ker(\varphi)$ that are all this cosets if we denote them $\overline e$ that is the $\ker(\varphi)$ itself then $\overline a, \overline b,...,\overline n$. $\ker(\varphi)$ must be a normal \ subgroup then all cosets are split into exactly the same number of elements that are contained in $\ker(\varphi)$. Also we know that $(\ast) \ \ker(\varphi)=\{g\in G\ | \ (\varphi(g)=e_\dot G$) in other words every element from the $\ker(\varphi)$ must map a element from his domain to the co domain in $\dot G$ and that should be the identity element. Now my confusion comes from this the theorem says that all the elements in the same coset are mapped to the same element in $\dot G$. The proof is that if we take an element from the coset, for example, $\overline a$ it should be in the following form: $$\overline a \ \circ n \ where: \\n\in \ker(\varphi) \\ a\in \overline a$$ so if we use the theorem we get: $$\varphi(a \circ n ) =\varphi(a) \circ\varphi(n)$$ that sends the result of the composition in the domain into a result from the composition in the co domain if I am undesranding it right and $\varphi(n)$ is obviously the identity element because the theorem states $(\ast)$ that every g that is mapped from the domain to the co-domain from the kernel is onto the identity element and i have take n $n\in \ker(\varphi)$ then $$\varphi(a \circ n)= \varphi(a) \circ E_\dot g=\varphi(a)$$ this proofs that if we take two elements from our quotient group that they will end up mapped to the same element in the co-domain. And how this proofs that all elements in one coset will end up into just one element in the co-domain. Because we prove this by taking two elements and one is from the subgroup and the other from the coset, that means that these two elements are not both from the coset $\overline a$ itself or I am missing something? Thank you for any help in advance.

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Note that an element in the group $\dfrac{G}{\ker\varphi}$ is of the form $[a]=\{a\cdot n\mid n\in\ker\varphi\}$ for some $a\in G$. Now every element of $[a]$ gets mapped to the same element, say $\varphi(a)$. So $\varphi$ restricted to $[a]$ is a constant. Hence you can define a map $\tilde{\varphi}:\dfrac{G}{\ker\varphi}\to H$ by $\tilde{\varphi}([a])=\varphi(a)$, which is well defined.

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  • $\begingroup$ So $[a]$ is in this form because the coset is generated by the subgroup $ker(\varphi)$ and because this we can consider that this element $[a]$ that is in the form $a*n \in \overline a$ is just an element from only the $\overline a$ in other words the element (a * n) can be considered to be two elements from the coset $\overline a$ ? $\endgroup$ – Boris Borovski Jun 10 at 7:07
  • $\begingroup$ Thank you i think i missed how [a] should be defined when i was looking at the proof. I am not sure i get it but i think this give me some insight. $\endgroup$ – Boris Borovski Jun 10 at 7:18
  • $\begingroup$ Yes, $a\cdot n$ is an element of the coset $[a]$ for every $n\in\ker\varphi$. $\endgroup$ – Thomas Shelby Jun 10 at 7:18
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    $\begingroup$ Thank you a lot. It is clear now. $\endgroup$ – Boris Borovski Jun 10 at 7:21

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