First Isomorphism Theorem (longer version)

$\textbf{First Isomorphism Theorem}$ i.e. The Fundamental Theorem of Homomorphisms.

Statement: Given $\varphi: G \longrightarrow H$ a group homomorphism, then:
(i) $ker \varphi \unlhd G$
(ii) $G/ker\varphi \cong im\varphi$
Where $ker \varphi$ is the kernel of the homomorphism $\varphi$ and $im\varphi$ is the image of $\varphi$, which is contained in $H$.

$\textit{Proof:}$ For part (i), we take 2 approaches. First, we note that problem 3.1.1 in Dummit and Foote establishes that with the given condition that $\varphi: G \rightarrow H$ is a group homomorphism, then the pre-image or pullback of $E\leq{H}$ is a subgroup of $G$. In symbols, $\varphi^{-1}(E) \leq G$. Additionally, $\varphi^{-1}(E) \unlhd G$ whenever $E \unlhd H$. $<1_H> \unlhd H$ since $1_H H 1_H^{-1}=H$. $$\varphi^{-1}(<1_H>)=\{g\in{G}|\varphi(g)=1_H \} = ker\varphi \Longrightarrow ker\varphi \unlhd G$$ We may also proceed as follows: Let $g\in{G}$ and let $x\in{ker\varphi} \subseteq G$. Then, since $\varphi$ is a homomorphism, $$\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g^{-1})$$ $$=\varphi(g)\cdot1_H\cdot\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=\varphi(gg^{-1})=\varphi(1_G)=1_H$$
So, $gxg^{-1}\in{ker\varphi}$ which implies $\ker\varphi \unlhd G$ We have shown that $ker\varphi$ is invariant under conjugation, so it is thus a normal subgroup.

For part (ii), we need to construct an isomorphism between $G/ker\varphi$ and $ \ im\varphi$.
Define $\Phi: G/ker\varphi \longrightarrow im\varphi$ s.t. $\Phi(aK)=\varphi(a)$.
So, $\Phi$ maps left (could also choose right) cosets in $G/ker\varphi$ to elements in $H$ that are images of elements in $G$ under $\varphi$.
Note that $G/ker\varphi$ is indeed a quotient group since $ker\varphi \unlhd G$.
Our goal is to show that $\Phi$ is an isomorphism. First, we show that $\Phi$ is actually well-defined:
For convenience of notation , let $K=ker\varphi$. Now, choose equal representatives from $G/ker\varphi$. Let $aK=bK$. Now, we want to show that these representatives are mapped to the same element, namely that $\Phi(aK)=\Phi(bK)$. If $aK=bK,$ then by operations on cosets, $$aK=bK \Longrightarrow aK(bK)^{-1}=(aK)(b^{-1}K)=(ab^{-1})K=K$$ Since $ab^{-1}K=K$, this implies that $ab^{-1}\in{K}$. With $ab^{-1}$ in the kernel of $\varphi$ we get that $\varphi(ab^{-1})=1_H$. $\varphi(ab^{-1})=\varphi(a)\varphi(b^{-1})=1_H \Longrightarrow \varphi(a)=\varphi(b)$. By our construction, $\varphi(a)= \Phi(aK)$ and $\varphi(b)=\Phi(bK)$, thus $\Phi(aK)=\Phi(bK)$. So, $\Phi$ is well-defined.
Since each of the steps in showing that $\Phi$ is well-defined can be reversed, we immediately get that $\Phi$ is injective (one-to-one). This is because the statement that $\Phi$ is one-to-one is the converse of the statement that $\Phi$ is well-defined. Lets demonstrate this anyways for completeness:
Let $\Phi(aK)=\Phi(bK)$. By construction, this means $\varphi(a)=\varphi(b) \Longrightarrow\varphi(a)\varphi(b)^{-1}=1_H \Longrightarrow \varphi(ab^{-1})=1_H \Longrightarrow ab^{-1}\in{K} \Longrightarrow ab^{-1}K=1_G K \Longrightarrow aK=bK $. Hence, $\Phi$ is one-to-one.

To show that $\Phi$ is surjective (onto), we pick and element $y\in{im\varphi}$ and show that there is a corresponding element in $G/K$ that maps to $y$. Let $y=\varphi(x)\in{im\varphi}$. Then, $\Phi(xK)=\varphi(x)=y$. Thus, $\Phi$ is onto.
To show that $\Phi$ is a homomorphism, we just need to show that $\Phi$ is operation preserving. This is the case since $$\Phi(aKbK)=\Phi(abK)=\varphi(ab)=\varphi(a)\varphi(b)=\Phi(aK)\Phi(bK)$$
So, $\Phi$ is an isomorphism, which shows that $G/K \cong im\varphi$. $\quad \blacksquare$