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The FIT states that if $\phi: G \rightarrow G'$ is a homomorphism, then Im($\phi$) $\cong$ $G$/Ker($\phi$).

I'm trying to break down this theorem into "understandable bits." Ker($\phi$) is the set of all elements of $G$ that gets mapped to $0$; so, this would mean that $G$/Ker($\phi$) is equal to the left cosets of these zero-mapped elements.

Im($\phi$) is the image of the homomorphism; but, according to the FIT, the image is isomorphic to the left cosets of Ker($\phi$). Doesn't this mean that these two things are "equivalent" (due to the isomorphism)? Would this translate that the left cosets of Ker($\phi$) partition the image of $\phi$? If so, what does that exactly mean? Or am I not on the right track to understanding this theorem? Thank you for your help.

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  • $\begingroup$ The cosets of $\ker(\varphi)$ partition $G$, not $\operatorname{im}(\varphi)$. $\endgroup$ – Hurkyl Dec 8 '16 at 11:57
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The left cosets of $\ker\phi$ don't partition $\phi(G)$, because $\ker \phi \subset G$ while $\phi(G) \subset G'$. The left cosets of $\ker\phi$ and the elements of $\phi(G)$ are not literally equal. But the FHT tells you that $G/\ker\phi$ and $\phi(G)$ have the same group structure, and are thus essentially the same.

Here's a more illuminating way of thinking about it: If $x,y \in G$ are in the same coset of $\ker\phi$, then $\phi(x) = \phi(y)$. And if $\phi(x) = \phi(y)$, then $x$ and $y$ are in the same coset of $\ker\phi$. So there is a one-to-one correspondence between cosets of $\ker\phi$ and elements of $\phi(G)$.

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    $\begingroup$ So, looking at the FHT diagram, is the theorem saying that: You can map an element $x$ in $G$ straight to some element in $G'$, or you can first map $x$ to some coset of Ker($\phi$), and then that coset can be mapped to the same element $g'$ in $G'$? $\endgroup$ – Max Dec 8 '16 at 6:30
  • $\begingroup$ Yeah, the diagram is saying that those two things are the same. $\endgroup$ – Ethan Alwaise Dec 8 '16 at 6:38
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    $\begingroup$ Ah, it's much more clear now. So, from your "illuminating way," it's true because $x$ gets first mapped to some coset $x + N$. And, that coset gets mapped to some value $k$ in the image of $\phi$. If $y$ is in the same coset, then that means $y$ gets mapped to the same $k$ as $x$ does. Is this what you were saying? $\endgroup$ – Max Dec 8 '16 at 6:42
  • $\begingroup$ Yeah, that's what I was saying. $\endgroup$ – Ethan Alwaise Dec 8 '16 at 6:46
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We define an equivalence relation $ \sim$ on $G$ as follows:

$x \sim y$ iff $xy^{-1} \in \ker( \phi)$.

We denote the equivalence class of $x \in G$ by $[x]$. Then define $T: G/ \ker( \phi) \to Im( \phi)$ by $T([x])=\phi(x)$.

Show that $T$ is well defined.

Are you now able to complete the proof ?

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