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.
Best Answer
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 $\phi(G)$.