Group Isomorphism
Forfatter
Srishti Patel
Sidst opdateret
5 år siden
Licens
Creative Commons CC BY 4.0
Resumé
Group Isomorphisms
Group Isomorphisms
\documentclass{beamer}
\usetheme{Boadilla}
\title{Group Isomorphisms}
\author{Srishti Patel(Roll no 23)}
\institute{Delhi Technological University}
\date{October 2019}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\begin{block}{Definition}
An isomorphism $\Phi$ from a group G to a group G' is one-one ,onto mapping that preseves the group operation,i.e
$\Phi(xy)=\Phi(x)\cdot\Phi(y)$
\end{block}
\textbf{Notation:} If G is isomorphic to G' then we denote it by
$G \cong G'$
\end{frame}
\begin{frame}[allowframebreaks]{Examples of Isomorphisms}
\begin{enumerate}
\item Let $G=(\mathbb{R},+)$ and $G'=(\mathbb{R^{+}},\cdot)$ be two groups then
$\Phi:G \rightarrow G'$ as $\Phi(x)=2^{x}$ is an isomorphism
$\bullet$ $\Phi$ is one-one -
Let $ \Phi(x)=\Phi(y)$
$ \Rightarrow 2^{x}=2^{y}$
$ \Rightarrow 2^{x-y}=1$
$\Rightarrow x-y=0$
$\Rightarrow x=y$
$\bullet$ $ \Phi$ is onto -
Let y $\in \mathbb{R^{+}}$, then $\exists x=\log_{2}y $ such that $\Phi(x)=2^{x}=2^{\log_{2}y}=y$
therefore, $\Phi$ is onto
$\bullet$ $\Phi$ is homomorphism -
$\Phi(x+y)=2^{x+y}=2^{x}\cdot2^{y}=\Phi(x)\Phi(y)$
therefore $\Phi$ is an isomorphism.
\item Let $G=SL(2\mathbb{R})$
Define $\Phi:G \rightarrow G$ as
$\Phi(A)=MAM^{-1} \forall A\in G $ where M is a fixed $2\times2$ matrix in $SL(2,\mathbb{R})$
then $\Phi$ is an isomorphism
$\bullet$ $\phi$ is one one -
Let $\phi(A)=\phi(B)$
$\Rightarrow MAM^{-1}=MBM^{-1}$
$\Rightarrow A=B$ (by pre and post multiplying by $M^{-1}$ and $M^{1}$ respectively)
$\bullet$ $\phi$ is onto -
Let A $\in$ G.Then $\exists$ $M^{-1}A M\in G$ such that $\Phi(M^{-1}AM)$$=M(M^{-1}AM)M^{-1}=A$
$\bullet$ $\phi$ is homomorphism-
$\Phi(A. B) = MABM^{-1}$
$=(MAM^{-1})MBM^{-1}$
$=\Phi(A).\Phi(B)$
Therefore $\Phi$ is an isomorphism called the \textbf{conjugation} by M
\item Let $G=\{1,\omega,\omega^{2}\}$,the group of cube roots of unity and
$G'=\{R_{0},R_{1},R_{2}\}$ the group of rotations in the plane through $0^{\circ}$,$120^{\circ}$ and $240^{\circ}$ respectively.
The mapping $\theta :G \rightarrow G'$ given by $\theta(1)=R_{0}$,$\theta(\omega)=R_{1}$ and $\theta(\omega^{2})=R_{2}$
defines an isomorphism of G onto G'.
\end{enumerate}
\end{frame}
\begin{frame}[allowframebreaks]{Properties of isomorphism}
Let $\theta:G\rightarrow G'$ be an isomorphism of G onto G'.Let e and e' be the identity elements of G and G' respectively.Then
\begin{enumerate}
\item $\theta(e)=e'$
\textbf{ proof:}
Let $\theta(e)=a' \in G'$.Then $a'=\theta(e)=\theta(e.e)=\theta(e)\cdot \theta(e)=a'.a'$.
Thus $a'a'=a'=a'e'$ ,by left cancellation law $a'=e'$.Hence $\theta(e)=e'$
\item $\theta(a^{-1})= \{\theta(a) \}^{-1}$ for all $a\in G$
\textbf{proof:}
$\theta(a^{-1})\cdot\theta(a)=\theta(a^{-1}.a)=\theta(e)=e' $ and $\theta(a)\cdot\theta(a^{-1})=\theta(a.a^{-1})=\theta(e)=e'$ .Hence by uniqueness of inverse in G',$\theta(a^{-1})$ is the inverse of $\theta(a)$
\textbf{Remark:}
in the above properties the result is valid even if $\theta$ is one-one and homomorphism .It need not be onto.
\item $\forall n\in \mathbb{Z}$ and $\forall a\in G ,\theta(a^{n})=[\theta(a)]^{n}$
\textbf{Proof:}
\textbf{Case1:}
When n=0 then $\theta(a^{0})=\theta(e)=e'$
,also $[\theta(a)]^{n}=[\theta(a)]^{0}=e'$
therefore , $\theta(a^{n})=[\theta(a)]^{n}$
\textbf{Case 2:}
When $n\in \mathbb{Z}^{+}$
$$ \theta(a^{n})=\theta(a\cdots a)\{ n times\}$$
$$=\theta(a)\cdot\theta(a)\cdots\theta(a) \{ n times\}$$
$$=[\theta(a)]^{n}$$
\textbf{Case 3:}
When $n\in \mathbb{Z}^{-}$ , Let $n=-m;m\in \mathbb{Z}$
$$ \theta(a^{n})=\theta(a^{-m})=[\theta(a^{-1})]^{m}= [\theta(a)]^{-m}$$
$$=[\theta(a)]^{n}$$
\end{enumerate}
\end{frame}
\begin{frame}[allowframebreaks]{Theorems}
\begin{theorem}
Let G and G' be isomorphic.If G is abelian ,so is G'
\end{theorem}
\textbf{Proof:}
Let $\theta:G\rightarrow G'$ be isomorphism of G onto G'.Let $a',b' \in G' $.Since $\theta$ is onto ,there exists $a\in G$ and $b\in G$ such that $\theta(a)=a'$ and $\theta(b)=b'$.Now $a'\cdot b'=\theta(a)\cdot\theta(b)=\theta(ab)=\theta(ba)$ (Since G is abelian)$=\theta(b)\cdot\theta(a)=b'a'$.
Thus G' is abelian.
\textbf{Remark 1:} The above theoram is also true if $\theta$ is an onto homomorphism.
\textbf{Remark 2:} If G is abelian and G' is non abelian then G and G' cannot be isomorphic.
\begin{theorem}
Any infinite cyclic group is isomorphic to $Z$ and any finite cyclic group of order n is isomorphic to $ Z _{n}$
\end{theorem}
\textbf{Proof:}
\textbf{Case1:} Let $G=<a>$ be an infinite cyclic group.Define $f:Z \rightarrow G$
given by $f(n)=a^{n}$
then $f(m+n)=a^{m+n}=a^{m}\cdot a^{n}=f(m)\cdot f(n) $.
Therefore f is homomorphism.Since all the powers are distinct in G therefore f is one-one.By definition it is onto .
Hence $G\cong Z$
\textbf{Case 2:} Let G be a finite cyclic group of order n. $G=<a>$ such that $0(a)=n$. Let $f:Z_{n}\rightarrow G$ is defined by $f(k')=a^{k}$
f is well defined.
$l'+m'=k' \Leftrightarrow l+m\equiv k(modn) where 0\le k\le n-1 $
$\Leftrightarrow (l+m-k) \mid n $
$\Leftrightarrow (l+m-k)= np $
$\Leftrightarrow a^{l+m-np}=a^{k}$
$\Leftrightarrow a^{l+m}.(a^{n})^{-p}=a^{k}$
$\Leftrightarrow a^{l+m}=a^{k}$
$f(l'+m')=f(k')=a^{k}=a^{l+m}=a^{l}.a^{m}=f(l').f(m')$
f is homomorphism.
$ l'\neq m' \Rightarrow l \neq m \Rightarrow a^{l}\neq a^{m} \Rightarrow f(l') \neq f(m')$
Therefore f is one-one. Clearly f is onto hence f is an isomorphism
$G \cong Z_{n}$
\begin{block}{Corollary1}
Any two cyclic groups of the same order are isomorphic
\end{block}
\textbf{Proof:} \textbf{Case 1:}Let G and G' be finite cyclic groups of order n ,then $G \cong Z_{n} $ and $G' \cong Z_{n}$
(by the above theorem)therefore, $G\cong G'$
\textbf{Case 2:} Let G and G' be infinite cyclic groups.By previous theorem $G \cong Z$ and $G' \cong Z$ therefore, $G\cong G'$
\begin{block}{Remark}
For each prime p,there exists only one group(upto isomorphism) of order p i.e the cyclic group of order p
\end{block}
\end{frame}
\begin{frame}{First theorem of Isomorphism}
\begin{theorem}
Let $f:G\rightarrow G' $ be a homomorphism of G onto G'and kernel of f is K then
$\bullet $ K is normal in G
$\bullet \frac{G}{K} \cong G'$
\end{theorem}
\textbf{Proof:} Let $f': \frac{G}{K} \rightarrow G'$ be defined as $f'(aK)=f(a) $ for $a \in G$
Let $aK=bK \Leftrightarrow a^{-1}b \in K $
$\Leftrightarrow f(a^{-1}b)=e' $
$\Leftrightarrow f(a^{-1})\cdot f(b)=e' $
$\Leftrightarrow f(a^{-1})\cdot f(b)=e' $
$\Leftrightarrow f(a)=f(b) $
$\Leftrightarrow f'(aK)=f'(bK) $
Therefore f is well defined and one-one
Since f is onto hence f' is also onto
$f'(aKbK)=f'(abK)=f(ab)=f(a).(b) = f'(aK). f'(bK)$
Therefore f' is homomorphism and since f' is one-one and onto as well hence f is isomorphism
$$ \frac{G}{K} \cong G' $$
\end{frame}
\begin{frame}
\begin{block}{Lemma}
Let $f:G \rightarrow G'$ be a homomorphism then,
\begin{itemize}
\item if $H<G$ ,then $f(H)=H'<G'$
\item if H is normal in G anf f is onto then H' is normal in G'
\item if $ H'<G' \Rightarrow f^{-1}(H')=H < G$
\item if H' is normal in G'$\Rightarrow$ H is normal in G
,further if f is onto then $\frac{G}{H} \cong \frac{G'}{H'}$
\end{itemize}
\end{block}
\end{frame}
\begin{frame}{Second Theorem of Isomorphism}
\begin{theorem}
Let H and K be normal in G such that $K\subset H$ then
\begin{itemize}
\item $\frac{H}{K} \triangleleft \frac{G}{K}$
\item $\frac{G/K}{H/K} \cong \frac{G}{H}$
\end{itemize}
\end{theorem}
\textbf{Proof:}
Consider the projection map
$$ p:G \rightarrow \frac{G}{K} =G' $$
by $p(a)=aK$ where $a\in G.$
Since H is normal in G ,$p(H)=\frac{H}{K}=H'$
Consider $$ G/K=\{aK | a\in G \} $$
$$ H/K=\{aK | a\in H \} $$
$$ H'\triangleleft G'$$
also from previous lemma, we have
$ \frac{G'}{H'} \cong \frac{G}{H} $ that is
$\frac{G/K}{H/K} \cong \frac{G}{H}$
\end{frame}
\begin{frame}[allowframebreaks]{Third theorem of Isomorphism }
\begin{theorem}
Let $H,K < G$ with K is normal in G
\begin{itemize}
\item $ H\cap K$ is normal in H
\item $ \frac{H}{H \cap K} \cong \frac{HK}{K} $
\end{itemize}
\end{theorem}
\textbf{Proof:}
Since K is normal in G and $K\leq G$ therefore HK is a subgroup of G $\Rightarrow HK=KH.$
As K is normal in G and $HK \leq G$ thus K is normal in HK $\Rightarrow \frac{HK}{K}$ well defined.
Also $H \cap K$ is normal in H.
Let $x\in H\cap K$
$\Rightarrow x\in H $and $x\in K$, since $h\in H$ therefore $hxh^{-1} \in H$
also since K is normal in G and $x\in K$
therefore $hxh^{-1}\in K$ and hence $hxh^{-1}\in H\cap K$
define $ f:H\rightarrow \frac{HK}{K}$ by
$$ f(a)=aK $$
then $xK \in \frac{HK}{K}$ then $xK=(hk)K=hK=f(H)$
(for some $h\in H$ and $k\in K$)
Therefore f is onto
$$f(ab)=abK=aK.bK=f(a).f(b)$$
f is a homomorphism
$Kerf= \{a\in H |f(a)=K\}$
$= \{a\in H |aK=K\}$
$= \{a\in H |a\in K\}$
$=H\cap K$
From first isomorphism theorem we have
$$\frac{H}{H\cap K} \cong \frac{HK}{K} $$
\end{frame}
\begin{frame}[allowframebreaks]{Questions}
\begin{block}{Question 1}
Show that $<Q,+>$ cannot be isomorphic to $<Q^{*},\cdot> $ where $Q^{*}= Q-\{0\} $
\end{block}
\textbf{Solution:}
Suppose f is an isomorphism from Q to $Q^{*}$.Then $2\in Q^{*}$
,f is onto therefore,$\exists \alpha \in <Q^{*}>$ ,s.t. $f(\alpha)=2$
$$ \Rightarrow f(\frac{\alpha}{2}+\frac{\alpha}{2})=2 $$
$$ \Rightarrow f(\frac{\alpha}{2}).f(\frac{\alpha}{2})=2 $$
$\Rightarrow x^{2}=2$ where $x=f(\frac{\alpha}{2}) \in Q^{*}$
But that is a contradiction as there is no rational number x such that $x^{2}=2$.Hence the result follows
\begin{block}{Question 2}
Show that any finite cyclic group of order n is isomorphic to the quotient group $\frac{\mathbb{Z}}{N}$ where $<\mathbb{Z},+>$ is a group of integers and $N=<n>$
\end{block}
\textbf{Solution:}
Let $G=<a>$ be of order n
Define $f:\mathbb{Z} \rightarrow G$ s.t. $f(m)=a^{m}$
then f is clearly well defined and onto map.
Since $f(m+k)=a^{m+k}=a^{m}.a^{k}=f(m).f(k)$
f is a homomorphism and therefore by First theorem of Isomorphism,$G \cong \frac{\mathbb{Z}}{ker f}$
We show ker f$=N=<n>$
Now $m\in$ Ker f$=N=<n>$
$$\Leftrightarrow f(m)=e $$ $$\Leftrightarrow a^{m}=e $$
$$\Leftrightarrow O(a)|m $$
$$\Leftrightarrow n|m $$
$$\Leftrightarrow m\in <n>$$
Hence $G\cong \frac{\mathbb{Z}}{<n>}$
\begin{block}{Question 3}
If G is the additive group of reals and N is the subgroup of G consisting of integers ,prove that $\frac{G}{N}$ is isomorphic to the group H of all complex numbers of absolute value under multiplication.
\end{block}
\textbf{Solution:} Define a map
$f(\alpha)=e^{2\pi i\alpha}$
$|e^{2\pi i\alpha}|=|cos 2\pi\alpha+i sin 2\pi\alpha|$
$=\sqrt{cos^{2}(2\pi\alpha)+sin^{2}(2\pi\alpha)}=1$
We show f is onto
Let $h\in H$ be any element then $h=a+ib$
where $|a+ib|=1=\sqrt{a^{2}+{b}}$
$a+ib=r(cos\theta +isin\theta)$
$|(a+ib|=1\Rightarrow r=1$
$ a+ib=cos\theta +isin\theta=e^{i\theta}$
$f(\frac{\theta}{2\pi})=e^{\frac{\theta}{2\pi}\cdot 2\pi i}=e^{i\theta}$
$\Rightarrow \frac{\theta}{2\pi} $is the required pre image
Now we will show that f is a homomorphism as
$$ f(\theta_{1}+\theta_{2})=e^{2\pi (\theta_{1}+\theta_{2})i} $$
$$ =e^{2\pi \theta_{1}i}\cdot e^{2\pi \theta_{2}i}=f(\theta_{1})f(\theta_{2})$$
By first theorem of isomorphism $H\cong \frac{G}{ker f}$
We claim that ker f$=N$
Let $\alpha \in Ker f$
$$\Leftrightarrow f(\alpha)=1$$
$$\Leftrightarrow e^{2\pi i \alpha}=1$$
$$\Leftrightarrow cos2\pi\alpha+isin2\pi\alpha=1+i0$$
$$\Leftrightarrow cos2\pi\alpha=1,sin2\pi\alpha=0$$
$$\Leftrightarrow 2\pi\alpha=2\pi\alpha n \pm 0$$
$$\Leftrightarrow \alpha=n$$
$$\Leftrightarrow \alpha \in N $$
Hence Ker f $=N$
\end{frame}
\end{document}