A subgroup in Group theory
I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.
In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?
i.e.
If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?
group-theory
asked yesterday
bladiebla
233
add a comment |
I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.
In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?
i.e.
If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?
group-theory
asked yesterday
bladiebla
233
2
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday
add a comment |
I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.
In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?
i.e.
If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?
group-theory
asked yesterday
bladiebla
233
I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.
In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?
i.e.
If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?
group-theory
group-theory
asked yesterday
bladiebla
233
asked yesterday
bladiebla
233
asked yesterday
bladiebla
233
asked yesterday
bladiebla
233
asked yesterday
bladiebla
233
233
2
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday
add a comment |
2
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday
2
2
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday
add a comment |
3 Answers
3
active
oldest
votes
It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.
answered yesterday
Wuestenfux
3,6861411
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
add a comment |
No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
|
show 3 more comments
If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.
answered yesterday
ncmathsadist
42.3k259102
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.
answered yesterday
Wuestenfux
3,6861411
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
add a comment |
It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.
answered yesterday
Wuestenfux
3,6861411
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
add a comment |
It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.
answered yesterday
Wuestenfux
3,6861411
It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.
answered yesterday
Wuestenfux
3,6861411
answered yesterday
Wuestenfux
3,6861411
answered yesterday
Wuestenfux
3,6861411
answered yesterday
Wuestenfux
3,6861411
3,6861411
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
add a comment |
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
– Ovi
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Surely, the ''subgroup criterion''.
– Wuestenfux
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
Yes ${}{}{}{}{}$
– Ovi
yesterday
add a comment |
No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
|
show 3 more comments
No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
|
show 3 more comments
No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
pendermath
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
pendermath
15310
answered yesterday
pendermath
15310
15310
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
|
show 3 more comments
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
1
1
I don't see what this has to do with the question.
– Matt Samuel
yesterday
I don't see what this has to do with the question.
– Matt Samuel
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
– pendermath
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
– Matt Samuel
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
I am not assuming that is a subset of a group. Maybe I missed the question.
– pendermath
yesterday
|
show 3 more comments
If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.
answered yesterday
ncmathsadist
42.3k259102
add a comment |
If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.
answered yesterday
ncmathsadist
42.3k259102
add a comment |
If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.
answered yesterday
ncmathsadist
42.3k259102
If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.
answered yesterday
ncmathsadist
42.3k259102
answered yesterday
ncmathsadist
42.3k259102
answered yesterday
ncmathsadist
42.3k259102
answered yesterday
ncmathsadist
42.3k259102
42.3k259102
add a comment |
add a comment |
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2
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
– Axion004
yesterday