The statement is the First Isomorphism Theorem for Groups from Abstract Algebra by Dummit and Foote. This proof was left as a exercise, so I'd like to check if all is ok. In particular, I'm a bit worried about the (*) line. It looks a bit awkward, but is it too bad? Any suggestions?
Theorem: If $\varphi : G \rightarrow H$ is a homomorphism of groups, then $ker \varphi \trianglelefteq G$ and $G/ker \varphi \cong \varphi (G)$.
Proof: Let $\varphi:G \rightarrow H$ be a group homomorphism. If $g \in G$ and $h \in ker \varphi$, then we have that $ghg^{-1} \in ker \varphi$ because $$\varphi (ghg^{-1})= \varphi (g) \varphi (h) \varphi (g^{-1})= \varphi (g) \varphi (g^{-1}) = \varphi (gg^{-1}) = \varphi (1) =1,$$ so that $ker \varphi \trianglelefteq G$. Now, define $\phi: G/ker \varphi \rightarrow \varphi (G)$ by $$\phi (g(ker \varphi))=\varphi(g),$$ for each $g(ker \varphi) \in G/ker \varphi$, for some $g \in G$. This is well defined because if $g' \in g(ker \varphi),$ then $$\phi (g'(ker \varphi))= \varphi(g')= \varphi(g)= \phi (g(ker \varphi)).$$ Also, this is a group isomorphism because for each $\varphi (h) \in \varphi (G),$ for some $h \in G$, we have $$(*) \ \phi^{-1}\{\varphi(h)\}= \phi^{-1} \varphi[g(ker \varphi)]= \phi^{-1} \varphi(\pi^{-1}\{g(ker \varphi)\})= \{g(ker \varphi)\},$$ a set with a single element of $G/ker \varphi$, so that it is a bijection, and for every $g(ker \varphi), g'(ker \varphi) \in G/ker \varphi$, for some $g,g' \in G$, we have $$\phi [g(ker \varphi) g'(ker \varphi)]=\phi [(gg')(ker \varphi)]=\varphi(gg')=\varphi(g) \varphi(g')=\phi [g(ker \varphi)] \phi [g'(ker \varphi)],$$ so that it is a group homomorphism.
Best Answer
Your proof of normality, definition of $\phi$ and proof that $\phi$ is well-defined are spot on (though I'd personally write something like "let $H = \ker(\varphi)$" at the start to avoid writing $\ker(\varphi)$ everywhere, as it makes things look slightly cleaner). Also, your proof that the map is a homomorphism is correct.
As for your starred line, as you're showing injectivity, I'd suggest replacing it with a simpler argument. In particular, as you have a homomorphism, to show injectivity it suffices to show that the kernel is trivial. But if $gH \in \ker(\phi)$, then by definition $\varphi(g) = 1$, so $g \in H$ and $gH = H$, which is the identity element in the quotient group.
An alternative proof - which implicitly proves the claim about trivial kernel that I mentioned above - is to say that if $\phi(gH) = \phi(hH)$, then $\varphi(g) = \varphi(h)$, so that $\varphi(gh^{-1}) = 1$ and $gh^{-1} \in H$. Thus $gH = hH$, as required.
I don't entirely follow your argument at $(*)$; what is the $\pi$ that crops up in the penultimate expression? Also, do you mean to write $\phi^{-1}\varphi[h\ker(\varphi)]$ in the second expression? My guess is that $\pi$ is the projection map from $G$ to $G/\ker(\varphi)$, and that in the end you're slightly assuming what you're trying to prove.
As an example, of a case where this might not prove injectivity, let's suppose that $J$ and $H$ are normal subgroups of $G$ with $J < H$ (strict containment). Then if $H$ is the kernel of some homomorphism $\varphi$ out of $G$, then we consider the map $$\phi: G/J \longrightarrow \varphi(G)$$ given by $$\phi(gJ) = \varphi(g).$$ In particular, this isn't an injective map. Then following your line $(*)$, we say that $$\phi^{-1}(hJ) = \phi^{-1}\varphi[hJ] = \phi^{-1}\varphi(\pi^{-1}\{hJ\}),$$ where $\pi$ is the projection $G \rightarrow G/J$. Does your argument then prove injectivity in this case? As if it does, then there's a mistake somewhere!
(Alternatively, you might take $\pi$ to be the projection map $G \rightarrow G/\ker(\varphi) = G/H$. But in this case you can't work with it unless you know exactly what $\ker(\varphi)$ is, in which case you already know if it's injective or not!)