From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!
First isomorphism theorem on groups
Let $\varphi: G \to H$ be a group homomorphism. Then
$\ker(\varphi)$ is a normal subgroup of $G$
$\operatorname{im}(\varphi)$ is a subgroup of $H$
$G/\ker(\varphi) \cong \operatorname{im}(\varphi)$
My attempt:
If $g_1, g_2 \in \ker(\varphi)$, then $\varphi(g_1 g_2)=\varphi(g_1) \varphi(g_2) =e' e' =e'$. Moreover, $g \in \ker(\varphi) \implies$ $\varphi (g^{-1}) = (\varphi (g))^{-1} =(e')^{-1} =e' \implies g^{-1} \in \ker(\varphi) \implies \ker(\varphi)$ is a subgroup of $G$.
If $h \in g \ker(\varphi)$, then $h = g n$ for some $n \in G$ such that $\varphi(n) = e'$. Let $n' = g n g^{-1}$. Then $\varphi(n') = e'$ and $n' g = h \implies h \in \ker(\varphi) g \implies g \ker(\varphi) \subseteq \ker(\varphi) g$. Similarly, $\ker(\varphi) g \subseteq g \ker(\varphi) \implies g \ker(\varphi) = \ker(\varphi) g \implies \ker(\varphi)$ is normal. Assertion (1.) then follows.
If $h_1, h_2 \in \operatorname{im}(\varphi)$, then $h_1 = \varphi (g_1), h_2 = \varphi (g_2)$ for some $g_1, g_2 \in G \implies \varphi(g_1 g_2) =$ $\varphi(g_1) \varphi(g_2)= h_1 h_2 \implies h_1 h_2 \in \operatorname{im}(\varphi)$. If $h \in \operatorname{im}(\varphi)$, then $h = \varphi (g)$ for some $g \in G \implies$ $\varphi (g^{-1}) = (\varphi (g))^{-1} = h^{-1} \implies h^{-1} \in \operatorname{im}(\varphi) \implies \operatorname{im}(\varphi)$ is a subgroup of $H$. Assertion (2.) then follows.
Consider $$\begin {array}{lrcl} \psi : & G/\ker(\varphi) & \longrightarrow & \operatorname{im}(\varphi) \\ & g \ker(\varphi) & \longmapsto & \varphi(g) \end{array}$$
Let $g_1, g_2 \in G$ such that $g_1 \ker(\varphi) = g_2 \ker(\varphi)$. Then $(g_2)^{-1} g_1 \ker(\varphi) = \ker(\varphi) \implies$ $(g_2)^{-1} g_1 \in \ker(\varphi) \implies \varphi((g_2)^{-1} g_1) = e' \implies \varphi((g_2)^{-1}) \varphi(g_1) = e' \implies$ $(\varphi(g_2))^{-1} \varphi(g_1) = e' \implies \varphi(g_1) = \varphi(g_2) \implies \psi$ is well-defined. Conversely, $\varphi(g_1) = \varphi(g_2) \implies g_1 \ker(\varphi)$ $= g_2 \ker(\varphi) \implies \psi$ is injective. Clearly, $\psi$ is surjective.
I will use symbol $\cdot$ for the induced operation on $G/\ker(\varphi)$. Then $$\psi ((g_1 \ker(\varphi)) \cdot (g_2 \ker(\varphi))) = \psi ((g_1 g_2) \ker(\varphi)) = \varphi (g_1 g_2) = \varphi (g_1) \varphi (g_2) = \psi (g_1 \ker(\varphi)) \psi (g_2 \ker(\varphi))$$
So $\psi$ is a homomorphism.
