Group Isomorphism
Forfatter:
Srishti Patel
Sidst opdateret:
5 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
Group Isomorphisms
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
Group Isomorphisms
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\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}