Is it true that an isomorphism maps elements of the same order to each other?
$begingroup$
I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?
abstract-algebra group-theory
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
$endgroup$
add a comment |
$begingroup$
I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?
abstract-algebra group-theory
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
$endgroup$
$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
1
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11
add a comment |
$begingroup$
I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?
abstract-algebra group-theory
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
$endgroup$
I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
asked Nov 21 '16 at 20:42
PiccolManPiccolMan
265110
265110
$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
1
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11
add a comment |
$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
1
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11
$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
1
1
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
$endgroup$
add a comment |
$begingroup$
Yes,
because for isomorphism it must hold that $f( u times v) = u cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$
$a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$
Note
$f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
$endgroup$
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
add a comment |
$begingroup$
If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
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oldest
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$begingroup$
"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
$endgroup$
add a comment |
$begingroup$
"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
$endgroup$
add a comment |
$begingroup$
"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
$endgroup$
"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
answered Nov 21 '16 at 21:10
user247327user247327
10.6k1515
10.6k1515
add a comment |
add a comment |
$begingroup$
Yes,
because for isomorphism it must hold that $f( u times v) = u cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$
$a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$
Note
$f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
$endgroup$
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
add a comment |
$begingroup$
Yes,
because for isomorphism it must hold that $f( u times v) = u cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$
$a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$
Note
$f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
$endgroup$
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
add a comment |
$begingroup$
Yes,
because for isomorphism it must hold that $f( u times v) = u cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$
$a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$
Note
$f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
$endgroup$
Yes,
because for isomorphism it must hold that $f( u times v) = u cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$
$a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$
Note
$f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
edited Nov 21 '16 at 21:10
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
answered Nov 21 '16 at 20:49
cubeceptioncubeception
1189
1189
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
add a comment |
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
$begingroup$
Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:50
3
3
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
$endgroup$
– Stahl
Nov 21 '16 at 20:54
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
$begingroup$
@PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
$endgroup$
– cubeception
Nov 21 '16 at 21:00
1
1
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
$begingroup$
@Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
$endgroup$
– cubeception
Nov 21 '16 at 21:01
add a comment |
$begingroup$
If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
$endgroup$
add a comment |
$begingroup$
If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
$endgroup$
add a comment |
$begingroup$
If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
$endgroup$
If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
answered Apr 21 '18 at 17:25
M. NestorM. Nestor
776113
776113
add a comment |
add a comment |
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$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44
$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45
1
$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46
$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11