Mixing
Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$.
$begingroup$
Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 implies yxz =1$. If we take $G$ to be the group of $2times 2$ matrices and let $x = left( begin{array} { c c } { 1 } & { 2 } \ { 0 } & { 2 } end{array} right)$, $y = left( begin{array} { l l } { 0 } & { 1 } \ { 2 } & { 1 } end{array} right)$ and $z = left( begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \ { 1 } & { - 1 } end{array} right)$. Then $x y z = left( begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right) = 1$ but $y x z = left( begin{array} { c c } { 2 } & { - 2 } \ { 5 } & { - 9 / 2 } end{array} right) neq 1$.
I don't understand where my proof went wrong.
abstract-algebra group-theory
asked Jan 11 at 14:53
LazarusLazarus
1127
$endgroup$
add a comment |
$begingroup$
Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 implies yxz =1$. If we take $G$ to be the group of $2times 2$ matrices and let $x = left( begin{array} { c c } { 1 } & { 2 } \ { 0 } & { 2 } end{array} right)$, $y = left( begin{array} { l l } { 0 } & { 1 } \ { 2 } & { 1 } end{array} right)$ and $z = left( begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \ { 1 } & { - 1 } end{array} right)$. Then $x y z = left( begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right) = 1$ but $y x z = left( begin{array} { c c } { 2 } & { - 2 } \ { 5 } & { - 9 / 2 } end{array} right) neq 1$.
I don't understand where my proof went wrong.
abstract-algebra group-theory
asked Jan 11 at 14:53
LazarusLazarus
1127
$endgroup$
3
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00
add a comment |
$begingroup$
Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 implies yxz =1$. If we take $G$ to be the group of $2times 2$ matrices and let $x = left( begin{array} { c c } { 1 } & { 2 } \ { 0 } & { 2 } end{array} right)$, $y = left( begin{array} { l l } { 0 } & { 1 } \ { 2 } & { 1 } end{array} right)$ and $z = left( begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \ { 1 } & { - 1 } end{array} right)$. Then $x y z = left( begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right) = 1$ but $y x z = left( begin{array} { c c } { 2 } & { - 2 } \ { 5 } & { - 9 / 2 } end{array} right) neq 1$.
I don't understand where my proof went wrong.
abstract-algebra group-theory
asked Jan 11 at 14:53
LazarusLazarus
1127
$endgroup$
Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 implies yxz =1$. If we take $G$ to be the group of $2times 2$ matrices and let $x = left( begin{array} { c c } { 1 } & { 2 } \ { 0 } & { 2 } end{array} right)$, $y = left( begin{array} { l l } { 0 } & { 1 } \ { 2 } & { 1 } end{array} right)$ and $z = left( begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \ { 1 } & { - 1 } end{array} right)$. Then $x y z = left( begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right) = 1$ but $y x z = left( begin{array} { c c } { 2 } & { - 2 } \ { 5 } & { - 9 / 2 } end{array} right) neq 1$.
I don't understand where my proof went wrong.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 11 at 14:53
LazarusLazarus
1127
asked Jan 11 at 14:53
LazarusLazarus
1127
asked Jan 11 at 14:53
LazarusLazarus
1127
asked Jan 11 at 14:53
LazarusLazarus
1127
asked Jan 11 at 14:53
LazarusLazarus
1127
1127
3
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00
add a comment |
3
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00
3
3
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
edited Jan 11 at 15:00
pie314271
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
$endgroup$
add a comment |
$begingroup$
For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
$begingroup$
Since $G$ is a group then if $xin G$ , there exists $x^{-1}in G$ such that $$xcdot x^{-1}=x^{-1}cdot x=ein G$$having this result and exploiting the other properties of group we obtain $$x^{-1}cdot (xyz)=(x^{-1}cdot x)cdot yz=ecdot yz=yz=x^{-1}cdot e=x^{-1}$$therefore $$yzcdot x=yzx=x^{-1}cdot x=1$$and the statement has been proved.
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Here $xyz=1$ gives
$$begin{align}
x^{-1}xyz=x^{-1}1 & iff yz=x^{-1} \
&iff yzx=x^{-1}x \
& iff yzx=1,
end{align}$$
which gives
$$begin{align}
y^{-1}yzx=y^{-1}1 & iff zx=y^{-1} \
& iff zxy=y^{-1}y \
&iff zxy=1.
end{align}$$
edited Jan 11 at 19:25
Shaun
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
edited Jan 11 at 15:00
pie314271
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
edited Jan 11 at 15:00
pie314271
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
edited Jan 11 at 15:00
pie314271
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
$endgroup$
You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
edited Jan 11 at 15:00
pie314271
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
edited Jan 11 at 15:00
pie314271
2,114819
edited Jan 11 at 15:00
pie314271
2,114819
edited Jan 11 at 15:00
pie314271
2,114819
2,114819
answered Jan 11 at 14:59
Ben WBen W
2,276615
answered Jan 11 at 14:59
Ben WBen W
2,276615
answered Jan 11 at 14:59
Ben WBen W
2,276615
2,276615
add a comment |
add a comment |
$begingroup$
Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
$endgroup$
add a comment |
$begingroup$
Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
$endgroup$
add a comment |
$begingroup$
Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
$endgroup$
Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
answered Jan 11 at 14:55
WuestenfuxWuestenfux
4,5141413
4,5141413
add a comment |
add a comment |
$begingroup$
For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
$begingroup$
For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
$endgroup$
add a comment |
$begingroup$
For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
$endgroup$
For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
answered Jan 11 at 15:03
Michael HoppeMichael Hoppe
11k31836
11k31836
add a comment |
add a comment |
$begingroup$
Since $G$ is a group then if $xin G$ , there exists $x^{-1}in G$ such that $$xcdot x^{-1}=x^{-1}cdot x=ein G$$having this result and exploiting the other properties of group we obtain $$x^{-1}cdot (xyz)=(x^{-1}cdot x)cdot yz=ecdot yz=yz=x^{-1}cdot e=x^{-1}$$therefore $$yzcdot x=yzx=x^{-1}cdot x=1$$and the statement has been proved.
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Since $G$ is a group then if $xin G$ , there exists $x^{-1}in G$ such that $$xcdot x^{-1}=x^{-1}cdot x=ein G$$having this result and exploiting the other properties of group we obtain $$x^{-1}cdot (xyz)=(x^{-1}cdot x)cdot yz=ecdot yz=yz=x^{-1}cdot e=x^{-1}$$therefore $$yzcdot x=yzx=x^{-1}cdot x=1$$and the statement has been proved.
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Since $G$ is a group then if $xin G$ , there exists $x^{-1}in G$ such that $$xcdot x^{-1}=x^{-1}cdot x=ein G$$having this result and exploiting the other properties of group we obtain $$x^{-1}cdot (xyz)=(x^{-1}cdot x)cdot yz=ecdot yz=yz=x^{-1}cdot e=x^{-1}$$therefore $$yzcdot x=yzx=x^{-1}cdot x=1$$and the statement has been proved.
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
Since $G$ is a group then if $xin G$ , there exists $x^{-1}in G$ such that $$xcdot x^{-1}=x^{-1}cdot x=ein G$$having this result and exploiting the other properties of group we obtain $$x^{-1}cdot (xyz)=(x^{-1}cdot x)cdot yz=ecdot yz=yz=x^{-1}cdot e=x^{-1}$$therefore $$yzcdot x=yzx=x^{-1}cdot x=1$$and the statement has been proved.
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 11 at 15:01
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
add a comment |
add a comment |
$begingroup$
Here $xyz=1$ gives
$$begin{align}
x^{-1}xyz=x^{-1}1 & iff yz=x^{-1} \
&iff yzx=x^{-1}x \
& iff yzx=1,
end{align}$$
which gives
$$begin{align}
y^{-1}yzx=y^{-1}1 & iff zx=y^{-1} \
& iff zxy=y^{-1}y \
&iff zxy=1.
end{align}$$
edited Jan 11 at 19:25
Shaun
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
add a comment |
$begingroup$
Here $xyz=1$ gives
$$begin{align}
x^{-1}xyz=x^{-1}1 & iff yz=x^{-1} \
&iff yzx=x^{-1}x \
& iff yzx=1,
end{align}$$
which gives
$$begin{align}
y^{-1}yzx=y^{-1}1 & iff zx=y^{-1} \
& iff zxy=y^{-1}y \
&iff zxy=1.
end{align}$$
edited Jan 11 at 19:25
Shaun
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
add a comment |
$begingroup$
Here $xyz=1$ gives
$$begin{align}
x^{-1}xyz=x^{-1}1 & iff yz=x^{-1} \
&iff yzx=x^{-1}x \
& iff yzx=1,
end{align}$$
which gives
$$begin{align}
y^{-1}yzx=y^{-1}1 & iff zx=y^{-1} \
& iff zxy=y^{-1}y \
&iff zxy=1.
end{align}$$
edited Jan 11 at 19:25
Shaun
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
Here $xyz=1$ gives
$$begin{align}
x^{-1}xyz=x^{-1}1 & iff yz=x^{-1} \
&iff yzx=x^{-1}x \
& iff yzx=1,
end{align}$$
which gives
$$begin{align}
y^{-1}yzx=y^{-1}1 & iff zx=y^{-1} \
& iff zxy=y^{-1}y \
&iff zxy=1.
end{align}$$
edited Jan 11 at 19:25
Shaun
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
edited Jan 11 at 19:25
Shaun
9,083113683
edited Jan 11 at 19:25
Shaun
9,083113683
edited Jan 11 at 19:25
Shaun
9,083113683
9,083113683
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
answered Jan 11 at 16:43
Michael RozenbergMichael Rozenberg
102k1791195
102k1791195
add a comment |
add a comment |
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3
$begingroup$
I think you need to double-check your "By applying the same argument..." line. Are you sure it works out?
$endgroup$
– Adrian Keister
Jan 11 at 14:57
$begingroup$
Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other.
$endgroup$
– Adrian Keister
Jan 11 at 15:00