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.

  • $\begingroup$ The cosets of $\ker(\varphi)$ partition $G$, not $\operatorname{im}(\varphi)$. $\endgroup$ – Hurkyl Dec 8 '16 at 11:57

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 $\phi(G)$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 ?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.