What is the sufficient and necessary conditions that $-1 in G$, where $G$ is a multiplicative group of a...
$begingroup$
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
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
$endgroup$
add a comment |
$begingroup$
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
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
$endgroup$
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
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
edited Dec 31 '18 at 12:29
Shaun
10.9k113687
10.9k113687
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
asked Dec 31 '18 at 8:33
zongxiang yizongxiang yi
352110
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $2mid ord(G)$.
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
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 $(-1,-1)$.
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $2mid ord(G)$.
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
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 $2mid ord(G)$.
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
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 $2mid ord(G)$.
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
$endgroup$
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 $2mid ord(G)$.
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
answered Dec 31 '18 at 8:42
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
$begingroup$
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 $(-1,-1)$.
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
$endgroup$
add a comment |
$begingroup$
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 $(-1,-1)$.
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
$endgroup$
add a comment |
$begingroup$
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 $(-1,-1)$.
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
$endgroup$
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 $(-1,-1)$.
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
answered Dec 31 '18 at 8:38
A. PongráczA. Pongrácz
6,1171929
6,1171929
add a comment |
add a comment |
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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