Mixing
![Use xampp to allow external access from localhost. - I haven't a [# New XAMPP security concept] block](https://lh3.googleusercontent.com/-lgwpEEKxnro/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-hhgyfFXd4AGxoGKShq_X_LYjpIug/s72-c-mo/photo.jpg?sz=32)
Is multiplication as an operation available in groups, rings and fields over Z_p*?
$begingroup$
I've seen groups, rings, and fields described with a multiplication operation as well as a group defined as only having addition and subtraction (via inverse) operations. Is the reason the answer varies with respect to a group having or not having a multiplication operation dependent upon the type of numbers represented (e.g. integers, reals, etc) as well as the elements included (e.g. 0 is not in Z_p* because it doesn't have an inverse? Or do groups never have a multiplication operation?
abstract-algebra group-theory finite-groups
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
$endgroup$
|
show 4 more comments
$begingroup$
I've seen groups, rings, and fields described with a multiplication operation as well as a group defined as only having addition and subtraction (via inverse) operations. Is the reason the answer varies with respect to a group having or not having a multiplication operation dependent upon the type of numbers represented (e.g. integers, reals, etc) as well as the elements included (e.g. 0 is not in Z_p* because it doesn't have an inverse? Or do groups never have a multiplication operation?
abstract-algebra group-theory finite-groups
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
$endgroup$
3
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
1
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
1
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
1
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
1
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16
|
show 4 more comments
$begingroup$
I've seen groups, rings, and fields described with a multiplication operation as well as a group defined as only having addition and subtraction (via inverse) operations. Is the reason the answer varies with respect to a group having or not having a multiplication operation dependent upon the type of numbers represented (e.g. integers, reals, etc) as well as the elements included (e.g. 0 is not in Z_p* because it doesn't have an inverse? Or do groups never have a multiplication operation?
abstract-algebra group-theory finite-groups
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
$endgroup$
I've seen groups, rings, and fields described with a multiplication operation as well as a group defined as only having addition and subtraction (via inverse) operations. Is the reason the answer varies with respect to a group having or not having a multiplication operation dependent upon the type of numbers represented (e.g. integers, reals, etc) as well as the elements included (e.g. 0 is not in Z_p* because it doesn't have an inverse? Or do groups never have a multiplication operation?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
edited Jan 30 at 4:36
YuiTo Cheng
2,1863937
2,1863937
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
asked Jan 30 at 4:12
JohnGaltJohnGalt
1195
1195
3
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
1
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
1
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
1
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
1
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16
|
show 4 more comments
3
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
1
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
1
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
1
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
1
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16
3
3
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
1
1
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
1
1
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
1
1
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
1
1
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $circ$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $langle D,+,* rangle$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $ntimes n$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
$endgroup$
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
add a comment |
$begingroup$
Assuming $mathbb Z_p=mathbb Z/pmathbb Z$ (not the $p$-adic integers), and $mathbb Z_p^*$ is the group of units in $mathbb Z_p$, then multiplication exists as a valid operation and $(mathbb Z_p^*,times)$ is a group. Recall that the definition of a group is a set $G$ together with an operation $oplus$ such that $(G,oplus)$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $times$ over $mathbb Z$ is associative, and the map taking an integer $x$ to the residue class $[x]$ preserves this algebraic property. The similar argument works for inverses. We conclude that $(mathbb Z_p^*,times)$ is a group.
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $circ$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $langle D,+,* rangle$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $ntimes n$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
$endgroup$
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
add a comment |
$begingroup$
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $circ$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $langle D,+,* rangle$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $ntimes n$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
$endgroup$
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
add a comment |
$begingroup$
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $circ$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $langle D,+,* rangle$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $ntimes n$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
$endgroup$
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $circ$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $langle D,+,* rangle$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $ntimes n$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
edited Jan 30 at 22:58
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
answered Jan 30 at 7:20
hardmathhardmath
29.3k953101
29.3k953101
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
add a comment |
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
$begingroup$
What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation?
$endgroup$
– JohnGalt
Jan 30 at 17:16
1
1
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
$begingroup$
This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer.
$endgroup$
– JohnGalt
Jan 30 at 20:42
1
1
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
$begingroup$
@JohnGalt: Done!
$endgroup$
– hardmath
Jan 30 at 22:59
add a comment |
$begingroup$
Assuming $mathbb Z_p=mathbb Z/pmathbb Z$ (not the $p$-adic integers), and $mathbb Z_p^*$ is the group of units in $mathbb Z_p$, then multiplication exists as a valid operation and $(mathbb Z_p^*,times)$ is a group. Recall that the definition of a group is a set $G$ together with an operation $oplus$ such that $(G,oplus)$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $times$ over $mathbb Z$ is associative, and the map taking an integer $x$ to the residue class $[x]$ preserves this algebraic property. The similar argument works for inverses. We conclude that $(mathbb Z_p^*,times)$ is a group.
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
$endgroup$
add a comment |
$begingroup$
Assuming $mathbb Z_p=mathbb Z/pmathbb Z$ (not the $p$-adic integers), and $mathbb Z_p^*$ is the group of units in $mathbb Z_p$, then multiplication exists as a valid operation and $(mathbb Z_p^*,times)$ is a group. Recall that the definition of a group is a set $G$ together with an operation $oplus$ such that $(G,oplus)$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $times$ over $mathbb Z$ is associative, and the map taking an integer $x$ to the residue class $[x]$ preserves this algebraic property. The similar argument works for inverses. We conclude that $(mathbb Z_p^*,times)$ is a group.
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
$endgroup$
add a comment |
$begingroup$
Assuming $mathbb Z_p=mathbb Z/pmathbb Z$ (not the $p$-adic integers), and $mathbb Z_p^*$ is the group of units in $mathbb Z_p$, then multiplication exists as a valid operation and $(mathbb Z_p^*,times)$ is a group. Recall that the definition of a group is a set $G$ together with an operation $oplus$ such that $(G,oplus)$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $times$ over $mathbb Z$ is associative, and the map taking an integer $x$ to the residue class $[x]$ preserves this algebraic property. The similar argument works for inverses. We conclude that $(mathbb Z_p^*,times)$ is a group.
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
$endgroup$
Assuming $mathbb Z_p=mathbb Z/pmathbb Z$ (not the $p$-adic integers), and $mathbb Z_p^*$ is the group of units in $mathbb Z_p$, then multiplication exists as a valid operation and $(mathbb Z_p^*,times)$ is a group. Recall that the definition of a group is a set $G$ together with an operation $oplus$ such that $(G,oplus)$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $times$ over $mathbb Z$ is associative, and the map taking an integer $x$ to the residue class $[x]$ preserves this algebraic property. The similar argument works for inverses. We conclude that $(mathbb Z_p^*,times)$ is a group.
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
answered Jan 30 at 7:13
YiFanYiFan
4,9961727
4,9961727
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3
$begingroup$
Binary operations are abstract. They can be whatever you want (or need).
$endgroup$
– Randall
Jan 30 at 4:15
1
$begingroup$
Yes, $mathbb Z_{11}^*$ with multiplication is a group.
$endgroup$
– J. W. Tanner
Jan 30 at 4:38
1
$begingroup$
Yes, $mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field
$endgroup$
– J. W. Tanner
Jan 30 at 5:01
1
$begingroup$
Here, we are using $mathbb Z_{11}$ to mean $mathbb Z/ (11 mathbb Z ),$ i.e., integers modulo $11$.
$endgroup$
– J. W. Tanner
Jan 30 at 5:09
1
$begingroup$
$mathbb Z_{11}$ could mean p-adic integers
$endgroup$
– J. W. Tanner
Jan 30 at 5:16