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!
Second isomorphism theorem on groups
Let $G$ be a group, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$.
$SN$ is a subgroup of $G$
$S \cap N$ is a normal subgroup of $S$
$SN/N \cong S/(S \cap N)$
My attempt:
Let $s_1n_1, s_2n_2 \in SN$. Since $N$ is normal, $n_1s_2 = s_2n'_1$ for some $n'_1 \in N \implies s_1n_1s_2 n_2 = s_1s_2n'_1n_2 \in SN$. Let $sn \in SN$. Since $N$ is normal, $sn = n's$ for some $n' \in N \implies (sn)^{-1} = (n's)^{-1} = s^{-1}(n')^{-1} \in SN$. Assertion (1.) then follows.
Clearly, $S \cap N$ is a subgroup of $S$. Let $s\in S$. If $h \in s (S \cap N)$, then $h = sm$ for some $m \in S \cap N$. Let $m' = sms^{-1}$. Then $m' \in S$. Since $N$ is normal, $ms^{-1} = s^{-1} m''$ for some $m'' \in N \implies m' = ss^{-1} m'' = m'' \implies m' \in S \cap N \implies (S \cap N)s \ni m' s = h \implies$ $s(S \cap N) \subseteq (S \cap N)s$. Similarly, $(S \cap N)s \subseteq s(S \cap N)$. So $(S \cap N)s = s(S \cap N)$. Assertion (2.) then follows.
Clearly, $N$ is a normal subgroup of $SN$.
Approach 1:
Consider $$\begin {array}{lrcl} \psi : & S/(S \cap N) & \longrightarrow & SN/N\\ & s(S \cap N) & \longmapsto & sN \end{array}$$
Clearly, $s(S \cap N) \subseteq sN$. If $s_1(S \cap N) = s_2(S \cap N)$ then $\emptyset \neq s_1(S \cap N) \subseteq (s_1N \cap s_2N) \implies s_1N = s_2N \implies \psi$ is well-defined. Let $snN\in SN/N$. Then $snN=s(nN)=sN$ and thus $\psi$ is surjective.
We have $\psi (s_1(S \cap N)) = \psi (s_2(S \cap N)) \implies s_1N = s_2 N \implies (s_1)^{-1}s_2 \in N$ $\implies (s_1)^{-1}s_2 \in S \cap N$. Moreover, $s_1((s_1)^{-1}s_2)=s_2e$ where $e \in S \cap N$ is the identity element of $G \implies s_1(S \cap N) \cap s_2(S \cap N) \neq \emptyset \implies s_1(S \cap N) =$ $s_2(S \cap N) \implies \psi$ is injective.
We have $\psi (s_1(S \cap N) s_2(S \cap N)) = \psi (s_1s_2(S \cap N)) = s_1s_2N = (s_1N) (s_2N) =$ $\psi (s_1(S \cap N)) \psi (s_1(S \cap N))$. Hence $\psi$ is an isomorphism.
Approach 2:
Consider $$\begin {array}{lrcl} \psi : & S & \longrightarrow & SN/N\\ & s & \longmapsto & sN \end{array}$$
Clearly, $sN=seN \in SN/N$ for all $s \in S$. Similarly, $\psi$ is surjective. We have $\psi (s_1 s_2) = (s_1 s_2)N = (s_1N)(s_2N) = \psi(s_1) \psi(s_2)$. Hence $\psi$ is homorphism. We have $\ker(\psi) = \{s \in S \mid \psi(s)= N\} = \{s \in S \mid sN = N\}$. Clearly, $s \in N \implies$ $s \in \ker(\psi)$. If $s \notin N$ then $sN \cap N = \emptyset$. Hence $\ker(\psi) = S \cap N$.
By the first isomorphism theorem on groups, $S/\ker(\psi) = S/(S \cap N) \cong SN/N$.
