What is the sufficient and necessary conditions that $-1 in G$, where $G$ is a multiplicative group of a...












4












$begingroup$


I am trying to prove the following conjecture.




Let $(R, +,times)$ be a finite ring with an identity. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.




My attempt:



$Leftarrow$:
Case 1:If $text{Char}(R)=2$, then $-1=1$. Since $G$ is a multiplicative group, the indentity $1in G$. Hence $-1in G$.



Case 2: If $2mid d$, then ...(I do not know how to prove in this case.)



$Rightarrow$: Suppose that $-1 in G$. Then $(-1)^2=1in G$. The multiplicative order of $-1$ is either $2$ or $1$. If $text{Ord}(-1)=1$, then $-1=1$. Consequently, $text{Char}(R)=2$. If $text{Ord}(-1)=2$, then it must have $text{Ord}(-1) mid text{Ord}(G)$, i.e., $2mid d$.



===============



So does $2mid d$ implies that $-1in G$ ? If the conjecture does not hold, then can we have a sufficient and necessary condition of $-1 in G$?



Thanks to Arthur, we have the following result using the fact that $-1$ is the only element of order $2$.




Let $(R, +,times)$ be an integral domain. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.







abstract-algebra ring-theory finite-groups conjectures







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
    $endgroup$
    – zongxiang yi
    Dec 31 '18 at 8:43






  • 1




    $begingroup$
    Arthur's answer covers that elegantly.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:44










  • $begingroup$
    If I were you, I would accept Arthur's answer. That pretty much solves your problem.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:58
















4












$begingroup$


I am trying to prove the following conjecture.




Let $(R, +,times)$ be a finite ring with an identity. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.




My attempt:



$Leftarrow$:
Case 1:If $text{Char}(R)=2$, then $-1=1$. Since $G$ is a multiplicative group, the indentity $1in G$. Hence $-1in G$.



Case 2: If $2mid d$, then ...(I do not know how to prove in this case.)



$Rightarrow$: Suppose that $-1 in G$. Then $(-1)^2=1in G$. The multiplicative order of $-1$ is either $2$ or $1$. If $text{Ord}(-1)=1$, then $-1=1$. Consequently, $text{Char}(R)=2$. If $text{Ord}(-1)=2$, then it must have $text{Ord}(-1) mid text{Ord}(G)$, i.e., $2mid d$.



===============



So does $2mid d$ implies that $-1in G$ ? If the conjecture does not hold, then can we have a sufficient and necessary condition of $-1 in G$?



Thanks to Arthur, we have the following result using the fact that $-1$ is the only element of order $2$.




Let $(R, +,times)$ be an integral domain. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.







abstract-algebra ring-theory finite-groups conjectures







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
    $endgroup$
    – zongxiang yi
    Dec 31 '18 at 8:43






  • 1




    $begingroup$
    Arthur's answer covers that elegantly.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:44










  • $begingroup$
    If I were you, I would accept Arthur's answer. That pretty much solves your problem.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:58














4












4








4





$begingroup$


I am trying to prove the following conjecture.




Let $(R, +,times)$ be a finite ring with an identity. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.




My attempt:



$Leftarrow$:
Case 1:If $text{Char}(R)=2$, then $-1=1$. Since $G$ is a multiplicative group, the indentity $1in G$. Hence $-1in G$.



Case 2: If $2mid d$, then ...(I do not know how to prove in this case.)



$Rightarrow$: Suppose that $-1 in G$. Then $(-1)^2=1in G$. The multiplicative order of $-1$ is either $2$ or $1$. If $text{Ord}(-1)=1$, then $-1=1$. Consequently, $text{Char}(R)=2$. If $text{Ord}(-1)=2$, then it must have $text{Ord}(-1) mid text{Ord}(G)$, i.e., $2mid d$.



===============



So does $2mid d$ implies that $-1in G$ ? If the conjecture does not hold, then can we have a sufficient and necessary condition of $-1 in G$?



Thanks to Arthur, we have the following result using the fact that $-1$ is the only element of order $2$.




Let $(R, +,times)$ be an integral domain. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.







abstract-algebra ring-theory finite-groups conjectures







share|cite|improve this question











$endgroup$




I am trying to prove the following conjecture.




Let $(R, +,times)$ be a finite ring with an identity. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.




My attempt:



$Leftarrow$:
Case 1:If $text{Char}(R)=2$, then $-1=1$. Since $G$ is a multiplicative group, the indentity $1in G$. Hence $-1in G$.



Case 2: If $2mid d$, then ...(I do not know how to prove in this case.)



$Rightarrow$: Suppose that $-1 in G$. Then $(-1)^2=1in G$. The multiplicative order of $-1$ is either $2$ or $1$. If $text{Ord}(-1)=1$, then $-1=1$. Consequently, $text{Char}(R)=2$. If $text{Ord}(-1)=2$, then it must have $text{Ord}(-1) mid text{Ord}(G)$, i.e., $2mid d$.



===============



So does $2mid d$ implies that $-1in G$ ? If the conjecture does not hold, then can we have a sufficient and necessary condition of $-1 in G$?



Thanks to Arthur, we have the following result using the fact that $-1$ is the only element of order $2$.




Let $(R, +,times)$ be an integral domain. Let $G$ be a
subgroup of $(R,times)$ with order $d$. Then $-1in G$, if and only
if $2mid d$ or $text{Char}(R)=2$.







abstract-algebra ring-theory finite-groups conjectures




abstract-algebra ring-theory finite-groups conjectures






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share|cite|improve this question













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edited Dec 31 '18 at 12:29









Shaun

10.9k113687




10.9k113687










asked Dec 31 '18 at 8:33









zongxiang yizongxiang yi

352110




352110








  • 1




    $begingroup$
    @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
    $endgroup$
    – zongxiang yi
    Dec 31 '18 at 8:43






  • 1




    $begingroup$
    Arthur's answer covers that elegantly.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:44










  • $begingroup$
    If I were you, I would accept Arthur's answer. That pretty much solves your problem.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:58














  • 1




    $begingroup$
    @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
    $endgroup$
    – zongxiang yi
    Dec 31 '18 at 8:43






  • 1




    $begingroup$
    Arthur's answer covers that elegantly.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:44










  • $begingroup$
    If I were you, I would accept Arthur's answer. That pretty much solves your problem.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 8:58








1




1




$begingroup$
@A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
$endgroup$
– zongxiang yi
Dec 31 '18 at 8:43




$begingroup$
@A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ?
$endgroup$
– zongxiang yi
Dec 31 '18 at 8:43




1




1




$begingroup$
Arthur's answer covers that elegantly.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 8:44




$begingroup$
Arthur's answer covers that elegantly.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 8:44












$begingroup$
If I were you, I would accept Arthur's answer. That pretty much solves your problem.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 8:58




$begingroup$
If I were you, I would accept Arthur's answer. That pretty much solves your problem.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 8:58










2 Answers
2






active

oldest

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7












$begingroup$

Counterexample: let $R=Bbb Z_{12}$ with standard addition and multiplication, and $G={1,5}$.



As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2mid ord(G)$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    It is clearly not true: let $R:= (mathbb{Z}_3, +, cdot)times (mathbb{Z}_3, +, cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group ${(1,-1),(1,1)}$ does not contain $(-1,-1)$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Counterexample: let $R=Bbb Z_{12}$ with standard addition and multiplication, and $G={1,5}$.



      As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2mid ord(G)$.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        Counterexample: let $R=Bbb Z_{12}$ with standard addition and multiplication, and $G={1,5}$.



        As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2mid ord(G)$.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          Counterexample: let $R=Bbb Z_{12}$ with standard addition and multiplication, and $G={1,5}$.



          As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2mid ord(G)$.






          share|cite|improve this answer









          $endgroup$



          Counterexample: let $R=Bbb Z_{12}$ with standard addition and multiplication, and $G={1,5}$.



          As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2mid ord(G)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 8:42









          ArthurArthur

          123k7122211




          123k7122211























              4












              $begingroup$

              It is clearly not true: let $R:= (mathbb{Z}_3, +, cdot)times (mathbb{Z}_3, +, cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group ${(1,-1),(1,1)}$ does not contain $(-1,-1)$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                It is clearly not true: let $R:= (mathbb{Z}_3, +, cdot)times (mathbb{Z}_3, +, cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group ${(1,-1),(1,1)}$ does not contain $(-1,-1)$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  It is clearly not true: let $R:= (mathbb{Z}_3, +, cdot)times (mathbb{Z}_3, +, cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group ${(1,-1),(1,1)}$ does not contain $(-1,-1)$.






                  share|cite|improve this answer









                  $endgroup$



                  It is clearly not true: let $R:= (mathbb{Z}_3, +, cdot)times (mathbb{Z}_3, +, cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group ${(1,-1),(1,1)}$ does not contain $(-1,-1)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 8:38









                  A. PongráczA. Pongrácz

                  6,1171929




                  6,1171929






























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