Menu Close

Q-an-E-lementary-to-abstract-algebra-prove-that-the-order-of-an-element-in-quotient-group-Q-Z-is-finite-Notice-Q-Z-a-b-




Question Number 174879 by mnjuly1970 last updated on 13/Aug/22
     Q : (an  E lementary to abstract algebra )            prove that the  order  of              an  element  in” quotient group ” (Q , ⊕)/(Z , ⊕) is finite.            Notice:  (Q , ⊕)/(Z , ⊕) = { (a/b) + Z ∣  a,b ∈ Z , b ≠0 }                                 ≻ Source: John B .F raleigh book ≺
$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Q}}\::\:\left(\boldsymbol{{an}}\:\:\mathscr{E}\:\boldsymbol{{lementary}}\:\boldsymbol{{to}}\:\boldsymbol{{abstract}}\:\boldsymbol{{algebra}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\:\boldsymbol{{order}}\:\:\boldsymbol{{of}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{an}}\:\:\boldsymbol{{element}}\:\:\boldsymbol{{in}}''\:\boldsymbol{{quotient}}\:\boldsymbol{{group}}\:''\:\left(\mathbb{Q}\:,\:\oplus\right)/\left(\mathbb{Z}\:,\:\oplus\right)\:\boldsymbol{{is}}\:\boldsymbol{{finite}}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Notice}}:\:\:\left(\mathbb{Q}\:,\:\oplus\right)/\left(\mathbb{Z}\:,\:\oplus\right)\:=\:\left\{\:\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\:+\:\mathbb{Z}\:\mid\:\:\boldsymbol{{a}},\boldsymbol{{b}}\:\in\:\mathbb{Z}\:,\:\boldsymbol{{b}}\:\neq\mathrm{0}\:\right\}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succ\:\boldsymbol{{Source}}:\:\boldsymbol{{John}}\:\boldsymbol{{B}}\:.\boldsymbol{{F}}\:\boldsymbol{{raleigh}}\:\boldsymbol{{book}}\:\prec\:\:\:\: \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 13/Aug/22
let (a/b)+Z∈(Q/Z)  ;a,b∈Z,b≠0⇒  b∗((a/b)+Z)=((a/b)+Z)∗.._(b−times) . ∗((a/b)+Z)  =a+Z =_((a∈Z)) Z  since Z is the natural element  of (Q/Z) ,(a/b)+Z has finite order.
$${let}\:\frac{{a}}{{b}}+\mathbb{Z}\in\frac{{Q}}{\mathbb{Z}}\:\:;{a},{b}\in\mathbb{Z},{b}\neq\mathrm{0}\Rightarrow \\ $$$${b}\ast\left(\frac{{a}}{{b}}+\mathbb{Z}\right)=\left(\frac{{a}}{{b}}+\mathbb{Z}\right)\ast.\underset{{b}−{times}} {.}.\:\ast\left(\frac{{a}}{{b}}+\mathbb{Z}\right) \\ $$$$={a}+\mathbb{Z}\:\underset{\left({a}\in\mathbb{Z}\right)} {=}\mathbb{Z} \\ $$$${since}\:\mathbb{Z}\:{is}\:{the}\:{natural}\:{element} \\ $$$${of}\:\frac{{Q}}{\mathbb{Z}}\:,\frac{{a}}{{b}}+\mathbb{Z}\:{has}\:{finite}\:{order}. \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 13/Aug/22
mercey master..
$${mercey}\:{master}.. \\ $$
Commented by kaivan.ahmadi last updated on 13/Aug/22
thank you
$${thank}\:{you} \\ $$
Commented by kaivan.ahmadi last updated on 13/Aug/22
shoma irani bodin?
$${shoma}\:{irani}\:{bodin}? \\ $$
Commented by mnjuly1970 last updated on 14/Aug/22
 baleh ostad arjmand    eradatmandim
$$\:{baleh}\:{ostad}\:{arjmand} \\ $$$$\:\:{eradatmandim} \\ $$
Commented by kaivan.ahmadi last updated on 14/Aug/22
moafagh bashi. lotf dari man  ostad nistam.
$${moafagh}\:{bashi}.\:{lotf}\:{dari}\:{man} \\ $$$${ostad}\:{nistam}. \\ $$
Commented by mnjuly1970 last updated on 14/Aug/22
khahesh mikonam ghorban   bozorgvarid ...az wojood    jenabali estefadeh mikonim    zendeh bashid ...
$${khahesh}\:{mikonam}\:{ghorban} \\ $$$$\:{bozorgvarid}\:…{az}\:{wojood}\: \\ $$$$\:{jenabali}\:{estefadeh}\:{mikonim} \\ $$$$\:\:{zendeh}\:{bashid}\:… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *