Random math: 12

Source of most content: Gunhee Cho

1st isomorphism theorem

\[\text{Let } \varphi:G \rightarrow H \text{ be a group homomorphism. Then, } \bar{\varphi}:G/\ker \varphi \cong \text{Im}\varphi \ (g\ker \varphi \mapsto \varphi(g).\nonumber\]

Note that $\ker \varphi \trianglelefteq G \leadsto G/\ker \varphi$ is a group. In particular, $\bar{\varphi}$ is well-defined and is a group isomorphism.


  1. (Well-defined) Let $g \ker \varphi = g’ \ker \varphi \Leftrightarrow g(g’)^{-1} \in \ker \varphi \Leftrightarrow \varphi(g(g’)^{-1})=\varphi(g)\varphi(g’)^{-1}=e_H \Leftrightarrow \varphi(g) = \varphi(g’)$. Therefore, $\bar{\varphi}$ is well-defined.
  2. (homomorphism) Let $g \ker \varphi := \bar{g}$. Then, $\bar{\varphi}(\bar{g}\cdot \bar{g}’)= \bar{\varphi}(\bar{gg}’) (\because g\ker \varphi \cdot g’\ker \varphi = gg’ \ker \varphi)$. Then, $\bar{\varphi}(\bar{gg}’)=\varphi(\bar{gg}’)=\varphi(g)\varphi(g’)=\bar{\varphi}(\bar{g})\bar{\varphi}(\bar{g}’)$ since $\varphi$ is group homomorphism. Therefore, $\bar{\varphi}(\bar{g}\cdot \bar{g}’) = \bar{\varphi}(\bar{g})\bar{\varphi}(\bar{g}’)$, which implies $\bar{\varphi}$ is group homomorphism.
  3. (injective) If $\bar{\varphi}(\bar{g})=\bar{\varphi}(\bar{g}’)$, then $\varphi(g)=\varphi(g’) \leadsto \varphi(g(g’)^{-1})=e_H, \ g(g’)^{-1} \in \ker \varphi \Leftrightarrow g \ker \varphi = g’ \ker \varphi$.
  4. (surjective) Take any $\varphi(g) \in \text{Im} \varphi$. Then, we can find $\bar{g}$ s.t. $\bar{\varphi}(\bar{g})=\varphi(g)$. Therefore, $\bar{\varphi}$ is onto its $\text{Im}\varphi$.

e.g.) Let $G=(\mathbb{C}-\{0\}, \times)$ be a group and $\varphi: (\mathbb{R},+) \rightarrow S^1 := \{z \in \mathbb{C} \mid \mid z \mid = 1 \} \ (x \mapsto e^{2\pi i x})$. Since $\varphi(x+y) =$ $e^{2\pi i (x+y)}=e^{2\pi i x} \times e^{2\pi i y}=\varphi(x)\times \varphi(y)$, $\varphi$ is group homomorphism. And $\ker \varphi = \{ x\in \mathbb{R} \mid \varphi(x)=1 \} = $$ \{x \in \mathbb{R} \mid \cos 2\pi x + i \sin 2\pi y =1 \} =: \mathbb{Z}$. By 1st isomorphism thm, $\mathbb{R}/\mathbb{Z} \cong S^1$.


\[N \trianglelefteq G, f: G \rightarrow H \text{ be a group homomorphism. Then, } \bar{f}: G/N \rightarrow H \ (gN \mapsto f(g)) \text{ is well-defined iff } N \subseteq \ker f. \nonumber\]


  1. $(\Rightarrow)$ Let $g \in N$ then $gN=N \leadsto \bar{f}(gN)=f(g) =\bar{f}(N)=f(e)=e_H \leadsto g \in \ker f$. Therefore, $N \subseteq \ker f$.
  2. $(\Leftarrow)$ Let $N \subseteq \ker f$ and $g_1 N = g_2 N \Leftrightarrow g_1 g_2^{-1}\in N \subseteq \ker f \leadsto f(g_1 g_2^{-1})=e_H \Leftrightarrow f(g_1)=f(g_2)$ $\Leftrightarrow \bar{f}(g_1N)=\bar{f}(g_2N)$ by definition of $\bar{f}$.

\[\text{Gp homo. }f: G \rightarrow H, M \trianglelefteq G, K \trianglelefteq H. \text{ Then, } \bar{f}: G/M \rightarrow H/K (gM \mapsto f(g)K) \text{ is well-defined iff } f(M)\subseteq K. \nonumber\]


$\bar{f}$ is well-defined iff $g_1M=g_2M \leadsto f(g_1)K=f(g_2)K \Leftrightarrow g_1 g_2^{-1} \in M \leadsto f(g_1)f(g_2)^{-1}\in K$ $\Leftrightarrow g_1g_2^{-1} \in M \leadsto f(g_1g_2^{-1})\in K \Leftrightarrow f(M)\subseteq K.$

2nd isomorphism theorem

\[\text{Let } H \trianglelefteq G \text{ and } K \trianglelefteq G. \text{ Then, } H/H\cap K \cong HK/K. \nonumber\]


To apply 1st isomorphism theorem, consider two group homomorphisms (inclusion map and projection map) $i: H \rightarrow G, \ \pi : G \rightarrow G/K$. Then $\pi \circ i : H \stackrel{i}{\rightarrow} G \stackrel{\pi}{\rightarrow} G/K$ is a group homomorphism. Thus, $\ker (\pi \circ i)= \{ h \in H \mid hK = K\} = H \cap K \trianglelefteq H$ (Note that identity of $G/K$ is $K$). Then, by 1st isomorphism theorem, $H/H\cap K \cong \text{Im}(\pi \circ i)=\{hK \mid h \in H \} = \{hkK \mid h \in H, k \in K \} = HK/K$, i.e., $H/H\cap K \cong HK/K$.

3rd isomorphism theorem

\[H,K \trianglelefteq G, K \leqslant H (\leftrightarrow K \trianglelefteq H). \text{ Then, } (G/K)/(H/K) \cong G/H. \nonumber\]


Consider $Id_G: G \rightarrow G$. Then, by above proposition, $\overline{Id}:G/K \rightarrow G/H \ (gK \mapsto gH)$ is well-defined iff $Id(K) \subseteq K \Leftrightarrow K \subseteq H$, which is true by assumption. Since $K \leqslant H$, $\overline{Id}$ is surjective $(\text{Im}\overline{Id}=G/H)$ and $\ker (\overline{Id}) = \{ gK \mid gH=H \} = \{gK \in G/K \mid g \in H \} = H/K$. Therefore, by 1st isomorphism theorem, $(G/K)/(H/K) \cong G/H$.