Putnam: Understanding a proof that a cancellative semigroup $S$ s.t. for any $sin S$ the set of powers of $s$...
$begingroup$
$B2.$ Let $S$ be a non-empty set with a binary operation (written like multiplication) such that:
(1) it is associative;
(2) $ab = ac$ implies $b = c$;
(3) $ba = ca$ implies $b = c$;
(4) for each element, the set of its powers is finite.
Is $S$ necessarily a group?
I understand the solution given:
Answer: yes.
Let a be any element. We show that for some n > 1 we have an = a. The set of its powers is finite, so for some r > s we have ar = as. If s = 1, we are done. If not, put b = as-1, then b ar-s+1 = b a, so we may cancel to get an = a with n = r - s + 1 > 1. Now put e = an-1. Then we have ea = ae = a.
Now take any b. We have a(eb) = (ae)b = ab, and cancelling gives eb = b. Similarly, (be)a = b(ea) = ba, so be = b. Hence e is an identity.
Also a has an inverse. If n - 1 = 1, then a = e, so a is its own inverse. If n - 1 > 1, then an-2 is its inverse.
Now if b is any other element, we may use the same argument to find another identity f and an element c such that cb = bc = f. But we have e = ef = f, so the identity is unique and c is an inverse for b.
The problem I am having is that they don't mention anything about the closure of such set $S$ under the operation.
For example, take $S = {1,2,3}$ under multiplication. I wouldn't say this is a group because $2times3$ is not in $S$ but it fills all the criteria of the problem. What am I misunderstanding about closure? Is it something so obvious they don't even need to mention it in the solution? Thanks.
finite-groups contest-math proof-explanation semigroups binary-operations
edited Nov 28 '18 at 23:44
Shaun
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
$endgroup$
add a comment |
$begingroup$
$B2.$ Let $S$ be a non-empty set with a binary operation (written like multiplication) such that:
(1) it is associative;
(2) $ab = ac$ implies $b = c$;
(3) $ba = ca$ implies $b = c$;
(4) for each element, the set of its powers is finite.
Is $S$ necessarily a group?
I understand the solution given:
Answer: yes.
Let a be any element. We show that for some n > 1 we have an = a. The set of its powers is finite, so for some r > s we have ar = as. If s = 1, we are done. If not, put b = as-1, then b ar-s+1 = b a, so we may cancel to get an = a with n = r - s + 1 > 1. Now put e = an-1. Then we have ea = ae = a.
Now take any b. We have a(eb) = (ae)b = ab, and cancelling gives eb = b. Similarly, (be)a = b(ea) = ba, so be = b. Hence e is an identity.
Also a has an inverse. If n - 1 = 1, then a = e, so a is its own inverse. If n - 1 > 1, then an-2 is its inverse.
Now if b is any other element, we may use the same argument to find another identity f and an element c such that cb = bc = f. But we have e = ef = f, so the identity is unique and c is an inverse for b.
The problem I am having is that they don't mention anything about the closure of such set $S$ under the operation.
For example, take $S = {1,2,3}$ under multiplication. I wouldn't say this is a group because $2times3$ is not in $S$ but it fills all the criteria of the problem. What am I misunderstanding about closure? Is it something so obvious they don't even need to mention it in the solution? Thanks.
finite-groups contest-math proof-explanation semigroups binary-operations
edited Nov 28 '18 at 23:44
Shaun
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
$endgroup$
$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
1
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48
add a comment |
$begingroup$
$B2.$ Let $S$ be a non-empty set with a binary operation (written like multiplication) such that:
(1) it is associative;
(2) $ab = ac$ implies $b = c$;
(3) $ba = ca$ implies $b = c$;
(4) for each element, the set of its powers is finite.
Is $S$ necessarily a group?
I understand the solution given:
Answer: yes.
Let a be any element. We show that for some n > 1 we have an = a. The set of its powers is finite, so for some r > s we have ar = as. If s = 1, we are done. If not, put b = as-1, then b ar-s+1 = b a, so we may cancel to get an = a with n = r - s + 1 > 1. Now put e = an-1. Then we have ea = ae = a.
Now take any b. We have a(eb) = (ae)b = ab, and cancelling gives eb = b. Similarly, (be)a = b(ea) = ba, so be = b. Hence e is an identity.
Also a has an inverse. If n - 1 = 1, then a = e, so a is its own inverse. If n - 1 > 1, then an-2 is its inverse.
Now if b is any other element, we may use the same argument to find another identity f and an element c such that cb = bc = f. But we have e = ef = f, so the identity is unique and c is an inverse for b.
The problem I am having is that they don't mention anything about the closure of such set $S$ under the operation.
For example, take $S = {1,2,3}$ under multiplication. I wouldn't say this is a group because $2times3$ is not in $S$ but it fills all the criteria of the problem. What am I misunderstanding about closure? Is it something so obvious they don't even need to mention it in the solution? Thanks.
finite-groups contest-math proof-explanation semigroups binary-operations
edited Nov 28 '18 at 23:44
Shaun
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
$endgroup$
$B2.$ Let $S$ be a non-empty set with a binary operation (written like multiplication) such that:
(1) it is associative;
(2) $ab = ac$ implies $b = c$;
(3) $ba = ca$ implies $b = c$;
(4) for each element, the set of its powers is finite.
Is $S$ necessarily a group?
I understand the solution given:
Answer: yes.
Let a be any element. We show that for some n > 1 we have an = a. The set of its powers is finite, so for some r > s we have ar = as. If s = 1, we are done. If not, put b = as-1, then b ar-s+1 = b a, so we may cancel to get an = a with n = r - s + 1 > 1. Now put e = an-1. Then we have ea = ae = a.
Now take any b. We have a(eb) = (ae)b = ab, and cancelling gives eb = b. Similarly, (be)a = b(ea) = ba, so be = b. Hence e is an identity.
Also a has an inverse. If n - 1 = 1, then a = e, so a is its own inverse. If n - 1 > 1, then an-2 is its inverse.
Now if b is any other element, we may use the same argument to find another identity f and an element c such that cb = bc = f. But we have e = ef = f, so the identity is unique and c is an inverse for b.
The problem I am having is that they don't mention anything about the closure of such set $S$ under the operation.
For example, take $S = {1,2,3}$ under multiplication. I wouldn't say this is a group because $2times3$ is not in $S$ but it fills all the criteria of the problem. What am I misunderstanding about closure? Is it something so obvious they don't even need to mention it in the solution? Thanks.
finite-groups contest-math proof-explanation semigroups binary-operations
finite-groups contest-math proof-explanation semigroups binary-operations
edited Nov 28 '18 at 23:44
Shaun
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
edited Nov 28 '18 at 23:44
Shaun
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
edited Nov 28 '18 at 23:44
Shaun
9,065113683
edited Nov 28 '18 at 23:44
Shaun
9,065113683
edited Nov 28 '18 at 23:44
Shaun
9,065113683
9,065113683
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
asked Jul 16 '17 at 1:36
mtheorylordmtheorylord
1,761627
1,761627
$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
1
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48
add a comment |
$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
1
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48
$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
1
1
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A "binary operation on a set $S$" is a map $S times S to S$. Multiplication on $S = {1,2,3}$ is a map $S times S to {1,2,3,4,6,9}$ not $to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $mu : S times S to S$ and I have some subset $T subset S$. Then $mu$ restricts to a map $T times T to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
$endgroup$
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
add a comment |
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$begingroup$
A "binary operation on a set $S$" is a map $S times S to S$. Multiplication on $S = {1,2,3}$ is a map $S times S to {1,2,3,4,6,9}$ not $to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $mu : S times S to S$ and I have some subset $T subset S$. Then $mu$ restricts to a map $T times T to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
$endgroup$
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
add a comment |
$begingroup$
A "binary operation on a set $S$" is a map $S times S to S$. Multiplication on $S = {1,2,3}$ is a map $S times S to {1,2,3,4,6,9}$ not $to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $mu : S times S to S$ and I have some subset $T subset S$. Then $mu$ restricts to a map $T times T to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
$endgroup$
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
add a comment |
$begingroup$
A "binary operation on a set $S$" is a map $S times S to S$. Multiplication on $S = {1,2,3}$ is a map $S times S to {1,2,3,4,6,9}$ not $to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $mu : S times S to S$ and I have some subset $T subset S$. Then $mu$ restricts to a map $T times T to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
$endgroup$
A "binary operation on a set $S$" is a map $S times S to S$. Multiplication on $S = {1,2,3}$ is a map $S times S to {1,2,3,4,6,9}$ not $to S$.
It doesn't make sense to ask whether or not a binary operation is closed. "Closure" comes up when dealing with subalgebras (sub-algebraic structures). That is, we have a binary operation $mu : S times S to S$ and I have some subset $T subset S$. Then $mu$ restricts to a map $T times T to S$ and the question of closure is whether or not we can restrict the codomain to $T$ as well.
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
answered Jul 16 '17 at 1:44
Trevor GunnTrevor Gunn
14.5k32046
14.5k32046
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
add a comment |
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
$begingroup$
Ok. I understand. Thanks.
$endgroup$
– mtheorylord
Jul 16 '17 at 1:46
add a comment |
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$begingroup$
The most "compact" definition of a group I've ever seen is a non-empty set $G$ with an associative operation $times:G^2to G $ such that for any $a,bin G$ there is a unique $cin G$ and a unique $din G$ such that $atimes c=b=dtimes a.$
$endgroup$
– DanielWainfleet
Jul 16 '17 at 1:53
1
$begingroup$
A "binary operation" by definition is closed. Multiplication is not a binary operation on {1,2,3} because it is not closed.
$endgroup$
– fleablood
Jul 16 '17 at 1:56
$begingroup$
@DanielWainfleet This might interest you. Also, a semigroup S is a group if and only if for every $ain S$ we have $aS=S=Sa$.
$endgroup$
– Shaun
Nov 28 '18 at 23:48