Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 24: $G/\text{ker}\hspace{0.17em}f\simeq \text{im}\hspace{0.17em}f$

Group homomorphisms are for comparing groups.

Let $G$ and $H$ be groups. A group homomorphism from $G$ to $H$ is a function $f:G\to H$ such that

(a)	If $g_1,g_2\in G$ then $f\left(g_1{g}_2\right)=f\left(g_1\right)f\left(g_2\right)\text{.}$

Let $f:G\to H$ be a group homomorphism. The kernel of $f$ is $$\text{ker}\hspace{0.17em}f=\{g\in G\hspace{0.17em}|\hspace{0.17em}f\left(g\right)=1\}\text{.}$$ The image of $g$ is $$\text{im}\hspace{0.17em}f=\left\{f\left(g\right)\hspace{0.17em}\right|\hspace{0.17em}g\in G\}\text{.}$$

Let $f:G\to H$ be a group homomorphism. Then $\text{ker}\hspace{0.17em}f$ is a normal subgroup of $G\text{.}$

		Proof of crucial point.
	To show: If $g\in G$ and $k\in \text{ker}\hspace{0.17em}f$ then $gk{g}^{-1}\in \text{ker}\hspace{0.17em}f\text{.}$ Assume $g\in G$ and $k\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $gk{g}^{-1}\in \text{ker}\hspace{0.17em}f\text{.}$ To show: $f\left(gk{g}^{-1}\right)=1\text{.}$ Since $k\in \text{ker}\hspace{0.17em}f,$ $f\left(k\right)=1\text{.}$ Since $f\left(g{g}^{-1}\right)=f\left(1\right)=1$ then $f\left(g^{-1}\right)=f{\left(g\right)}^{-1}\text{.}$ So $$\begin{array}{ccc}{\displaystyle f\left(gk{g}^{-1}\right)}& =& {\displaystyle f\left(g\right)f\left(k\right)f\left(g^{-1}\right)}\\ & =& {\displaystyle f\left(g\right)f\left(k\right)f{\left(g\right)}^{-1}}\\ & =& {\displaystyle f\left(g\right)·1·f{\left(g\right)}^{-1}}\\ & =& {\displaystyle 1\text{.}}\end{array}$$ $\square$

Let $f:G\to H$ be a group homomorphism. Let $K=\text{ker}\hspace{0.17em}f\text{.}$

(a)	The function $$\begin{array}{cccc}f\prime :& G& ⟶& \text{im}\hspace{0.17em}f\\ & g& ⟼& f\left(g\right)\end{array}$$ is a surjective group homomorphism.
(b)	The function $$\begin{array}{cccc}\hat{f}:& \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.& ⟶& H\\ & gK& ⟼& f\left(g\right)\end{array}$$ is a well defined injective group homomorphism.
(c)	The function $$\begin{array}{cccc}\hat{f}\prime :& \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.& ⟶& \text{im}\hspace{0.17em}f\\ & gK& ⟼& f\left(g\right)\end{array}$$ is a well defined bijective group homomorphism.

		Proof in order of most crucial points to least crucial.
	(ba) To show: The function $\hat{f}:\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\to H$ give by $\hat{f}\left(gK\right)=f\left(g\right)$ is well defined. (i.e. is a function). To show: If $g_{1}K=g_{2}K$ then $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)\text{.}$ Assume $g_{1}K=g_{2}K\text{.}$ Then $g_1\in g_{2}K\text{.}$ So there exists $k\in K$ such that $g_1=g_{2}k\text{.}$ To show: $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)\text{.}$ To show: $f\left(g_1\right)=f\left(g_2\right)\text{.}$ $$f\left(g_1\right)=f\left(g_{2}k\right)=f\left(g_2\right)f\left(k\right)=f\left(g_2\right)·1=f\left(g_2\right)\text{.}$$ (bb) To show: The function $\hat{f}:\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\to H$ given by $\hat{f}\left(gK\right)=f\left(g\right)$ is injective. To show: If $g_{1}K,g_{2}K\in \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$ and $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)$ then $g_{1}K=g_{2}K\text{.}$ Assume $g_{1}K,g_{2}K\in \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$ and $\hat{f}\left(g_{1}K\right)=\hat{f}\left(g_{2}K\right)\text{.}$ Then $f\left(g_1\right)=f\left(g_2\right)\text{.}$ To show: $g_{1}K=g_{2}K\text{.}$ To show: $g_1\in g_{2}K,$ since $\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$ partitions $G\text{.}$ To show: There exists $k\in K$ such that $g_1=g_{2}k\text{.}$ To show: $k=g_2^{-1}{g}_1$ is an element of $K=\text{ker}\hspace{0.17em}f\text{.}$ To show: $f\left(g_2^{-1}{g}_1\right)=1\text{.}$ $$f\left(g_2^{-1}{g}_1\right)=f\left(g_2^{-1}\right)f\left(g_1\right)=f{\left(g_2\right)}^{-1}f\left(g_1\right)=f{\left(g_2\right)}^{-1}f\left(g_2\right)=1\text{.}$$ $$⋮$$ $\square$

$$G={GL}_2\left(ℂ\right)=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a,b,c,d\in ℂ\hspace{0.17em}\text{and}\hspace{0.17em}ad-bc\ne 0\}\text{.}$$ Then $$\begin{array}{cccc}f:& {GL}_2\left(ℂ\right)& ⟶& {GL}_1\left(ℂ\right)\\ & A& ⟼& \text{det}\left(A\right)\end{array}$$ is a group homomorphism because $\text{det}\left(A_1{A}_2\right)=\text{det}\left(A_1\right)\text{det}\left(A_2\right)\text{.}$ Then $$\begin{array}{ccc}{\displaystyle \text{ker}\hspace{0.17em}f}& =& {\displaystyle \{A\in {GL}_2\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}f\left(A\right)=1\}}\\ & =& {\displaystyle \{A\in {GL}_2\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}\text{det}\left(A\right)=1\}}\\ & =& {\displaystyle {SL}_2\left(ℂ\right)\text{.}}\end{array}$$ Since $\text{im}\hspace{0.17em}f={ℂ}^{\times }$ (because $\text{det}\left(\begin{array}{cc}a& 0\\ 0& 1\end{array}\right)=a\text{)}$ then $$\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$\text{ker}\hspace{0.17em}f$}\right.⥲{ℂ}^{\times }\text{.}$$ So $$\raisebox{1ex}{${GL}_2\left(ℂ\right)$}\!\left/ \!\raisebox{-1ex}{${SL}_2\left(ℂ\right)$}\right.⥲{ℂ}^{\times }$$ and ${SL}_2\left(ℂ\right)$ is a normal subgroup of ${GL}_2\left(ℂ\right)\text{.}$

Notes and References

These are a typed copy of Lecture 24 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 16, 2011.

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