Basic properties of group homomorphism

 

Suppose $f:G\to G^{\prime }$ is a group homomorphism, and $e$ and $e^{\prime }$ are the identity elements of $G$ and $G^{\prime }$, respectively.

 

1. Identity preservation: $f(e)=e^{\prime }$

 

pf) $\mathrm{\forall }g\in G,f(g)=f(eg)=f(e)f(g)$. Thus $f(e)=e^{\prime }$.

 

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2. Inverse preservation: $\mathrm{\forall }g\in G,f(g^{-1})=f(g{)}^{-1}$

 

pf) $\mathrm{\forall }g\in G,e^{\prime }=f(e)=f(g{g}^{-1})=f(g)f(g^{-1})$. Thus $f(g{)}^{-1}=f(g^{-1})$.

 

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3. Kernel is a normal subgroup: $\ker f⊴G$

 

pf) $\mathrm{\forall }h\in \ker f,\mathrm{\forall }g\in G$,

 

$$\begin{array}{rl}f(gh{g}^{-1})& =f(g)f(h)f(g^{-1})\\ & =f(g)f(g^{-1})=e^{\prime }\end{array}$$

 

Thus $gh{g}^{-1}\in \ker f$.

 

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4. Image is a subgroup of G'.

 

pf) (closedness) $\mathrm{\forall }u,v\in f(G)$, let $f(x)=u,f(y)=v$. Then

 

$$\begin{array}{rl}uv& =f(x)f(y)\\ & =f(xy)\in f(G).\end{array}$$

 

(identity element) $e^{\prime }\in f(G)$ since $f(e)=e^{\prime }$.

 

(inverse element) Pick any $u\in f(G)$. Let $f(x)=u$. Then

 

$$e^{\prime }=f(e)=f(x{x}^{-1})=f(x)f(x^{-1})=uf(x^{-1})$$.

 

Thus $f(x^{-1})$ is the inverse of $u$.

 

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5. 1st isomorphism thm: $G/\ker f\cong f(G)$

 

pf) Define a function

 

$$\begin{array}{rl}f:G/\ker f& \to f(G)\\ g\ker f& \mapsto f(g)\end{array}$$

 

(well-definedness) If $g_1\ker f=g_2\ker f$ for $g_1,g_2\in G$, then

 

$$\begin{array}{rl}{g}_1^{-1}{g}_2& \in \ker f\\ f(g_1^{-1}{g}_2)& =e^{\prime }\\ f(g_1^{-1})f(g_2)& =e^{\prime }\\ f(g_1)& =f(g_2)\end{array}$$

 

(1-1) If $f(g_1)=f(g_2)$, then

 

$$\begin{array}{rl}f(g_1)& =f(g_2)\\ f(g_1^{-1}{g}_2)& =e^{\prime }\\ g_1^{-1}{g}_2& \in \ker f\\ g_1\ker f& =g_2\ker f\end{array}$$

 

(onto) $\mathrm{\forall }f(g)\in G^{\prime },\mathrm{\exists }g\ker f\in G/\ker f$.

 

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6. Image of a subgroup is a subgroup.

 

pf) Let $H$ be a subgrup of $G$.

 

(closedness) Pick any $f(x)$ and $f(y)$ from $f(G)$. Then $xy\in G$ and $f(x)f(y)=f(xy)\in f(G)$.

 

(identity element) Because $H\le G$, $e\in H$ and $e^{\prime }\in f(H)$.

 

(inverse element) Pick any $f(x)$ from $f(G)$. Because $x^{-1}\in G$, $f(x^{-1})=f(x{)}^{-1}\in f(G)$ which is the inverse of $f(x)$.

 

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7. Preimage of a subgroup is a subgroup.

 

pf) Let $H^{\prime }$ be a subgroup of $G^{\prime }$ and $H_0=f^{-1}(H^{\prime })$.

 

(closedness) Pick any $x,y\in H_0$. Because $f(x)\in H^{\prime }$, $f(y)\in H^{\prime }$ and $H^{\prime }$ is a subgroup of $G^{\prime }$, $f(xy)=f(x)f(y)\in H^{\prime }$. By definition of preimage, $xy\in H_0$.

 

(identity element) Because $e^{\prime }\in H^{\prime }$, $e\in \ker f\subseteq H_0$.

 

(inverse element) Pick any $x\in H_0$. Because both $f(x)$ and $f(x^{-1})$ are in $H^{\prime }$, $x^{-1}\in H_0$.

 

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8. $\ker f\subseteq H\le G\Rightarrow f^{-1}(f(H))=H$.

 

pf) $f^{-1}(f(H))\supseteq H$: Trivial.

 

$f^{-1}(f(H))\subseteq H$: Let $x\in f^{-1}(f(H))$ which means $f(x)\in f(H)$. Suppose $x\notin H$. Then $\mathrm{\exists }h\in H$ s.t. $f(x)=f(h)$. Because $f$ is a group homomorphism,

 

$$\begin{array}{rl}e& =f(h)f(h^{-1})\\ & =f(x)f(h^{-1})\\ & =f(x{h}^{-1})\end{array}$$

 

Therefore $x{h}^{-1}\in \ker f\subseteq H$ which contradicts with the assumption that $x\notin H$.