Theorem: First Isomorphism Theorem

Let G and H be groups

$\varphi :G\rightarrow H$ a Homomorphism

Then, $G/\ker(\varphi) \cong \varphi(G)$ via the Isomorphism

$$\Psi:G/\ker(\varphi)\rightarrow \varphi(G)$$

$$(\ker(\varphi)) g \mapsto (g)\varphi$$

The difficulty lies in showing that $\Psi$ is one-to-one.

To show that a map is one-to-one, we show that the only element that gets mapped to the identity is the identity itself.

For ease of notation, let $K=\ker(\varphi)$

$$\ker(\Psi)=\left \{ Kg \in G/K \mid e=\varphi (g)\right \}$$ $$=\left \{ Kg \in G/K \mid g \in \ker(\varphi) \right\} =K$$

Why is the defining property "$e=\varphi(g)$"?

Isn't the definition of $\ker(\Psi) \text{ really just } \ker(\Psi)= \left\{ Kg \in G/K \mid e=\Psi(Kg) \right\}?$

In the second line of the proof, how is $g \in \ker(\varphi)$?

Thanks in advance.

  • $\begingroup$ Why are you putting the function BEHIND things? it should be $\varphi(G)$ $\endgroup$ – Zelos Malum Apr 21 '16 at 4:29
  • $\begingroup$ @ZelosMalum It is the same thing.. $\endgroup$ – Mathematicing Apr 21 '16 at 4:49
  • $\begingroup$ Not necciserily, it may mean many things if it comes behind that has nothing to do with homomorphisms and it's a very strange and peculiar notation. Better sticking to the norm. $\endgroup$ – Zelos Malum Apr 21 '16 at 4:50
  • $\begingroup$ My professor uses this convention.. $\endgroup$ – Mathematicing Apr 21 '16 at 5:00
  • $\begingroup$ @ZelosMalum Many group theorists write mappings on the right, so that composition is from left to right. It's annoying at first, but you get used to it after a while. $\endgroup$ – Bungo Apr 21 '16 at 8:19

I'm not sure I understand this line in your post: $(\ker(\varphi)) gx \mapsto x^2 (g)\varphi$

But you are spot on in defining $\Psi:G/\ker(\varphi)\rightarrow \varphi(G)$ and by this definition, $\Psi$ maps an element $Kg$ to some $\varphi(g)$. If this $Kg \in ker(\Psi)$ as well, it goes to $e$, and hence, we have: $\varphi(g) = e$. Hence $g \in ker (\varphi)$.

  • $\begingroup$ @Mathematicing: It seems there are plenty of posts on the First Isomorphism Theorem already. Have you looked at those? Some of them might help. $\endgroup$ – Damodar Apr 21 '16 at 5:41
  • $\begingroup$ I've made the correction. Sorry for the blunder. $\endgroup$ – Mathematicing Apr 21 '16 at 8:20
  • $\begingroup$ It slipped my mind to ask 'how' then is $Ker(\Psi)=K?$ Sure, by definition it is but how do I make the justification? For this to be true, it must be that g=e. The fact that $g \in Ker(\varphi)$ implies that g is some element that is mapped to the identity under $\varphi$ but does not necessarily implies that g=e. $\endgroup$ – Mathematicing Apr 22 '16 at 4:13

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