A subgroup in Group theory
Clash Royale CLAN TAG#URR8PPP
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?
$endgroup$
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?
$endgroup$

2$begingroup$
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.
$endgroup$
– Axion004
Jan 5 at 14:29
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?
$endgroup$
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?

2$begingroup$
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.
$endgroup$
– Axion004
Jan 5 at 14:29
add a comment 

2$begingroup$
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.
$endgroup$
– Axion004
Jan 5 at 14:29
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.
$endgroup$
– Axion004
Jan 5 at 14:29
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.
$endgroup$
– Axion004
Jan 5 at 14:29
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) $e= gg^{1}in U$.
$endgroup$

$begingroup$
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41

$begingroup$
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42 
$begingroup$
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
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: $1=a(ba) neq (ab)a=a$
$endgroup$

1$begingroup$
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16 
$begingroup$
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.
$endgroup$
– pendermath
Jan 5 at 19:37

$begingroup$
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38 
$begingroup$
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39 
$begingroup$
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39

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 $G$.
$endgroup$
add a comment 
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3 Answers
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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) $e= gg^{1}in U$.
$endgroup$

$begingroup$
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41

$begingroup$
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42 
$begingroup$
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
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) $e= gg^{1}in U$.
$endgroup$

$begingroup$
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41

$begingroup$
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42 
$begingroup$
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
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) $e= gg^{1}in U$.
$endgroup$
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) $e= gg^{1}in U$.

$begingroup$
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41

$begingroup$
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42 
$begingroup$
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
add a comment 

$begingroup$
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41

$begingroup$
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42 
$begingroup$
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41
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 $ab$. And when you will study rings, you will switch to the additive notation for groups)
$endgroup$
– Ovi
Jan 5 at 14:41
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42
Surely, the ”subgroup criterion”.
$endgroup$
– Wuestenfux
Jan 5 at 14:42
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
Yes ${}{}{}{}{}$
$endgroup$
– Ovi
Jan 5 at 14:43
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: $1=a(ba) neq (ab)a=a$
$endgroup$

1$begingroup$
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16 
$begingroup$
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.
$endgroup$
– pendermath
Jan 5 at 19:37

$begingroup$
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38 
$begingroup$
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39 
$begingroup$
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39

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: $1=a(ba) neq (ab)a=a$
$endgroup$

1$begingroup$
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16 
$begingroup$
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.
$endgroup$
– pendermath
Jan 5 at 19:37

$begingroup$
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38 
$begingroup$
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39 
$begingroup$
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39

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: $1=a(ba) neq (ab)a=a$
$endgroup$
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: $1=a(ba) neq (ab)a=a$

1$begingroup$
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16 
$begingroup$
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.
$endgroup$
– pendermath
Jan 5 at 19:37

$begingroup$
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38 
$begingroup$
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39 
$begingroup$
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39

show 3 more comments

1$begingroup$
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16 
$begingroup$
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.
$endgroup$
– pendermath
Jan 5 at 19:37

$begingroup$
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38 
$begingroup$
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39 
$begingroup$
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16
I don’t see what this has to do with the question.
$endgroup$
– Matt Samuel
Jan 5 at 19:16
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.
$endgroup$
– pendermath
Jan 5 at 19:37
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.
$endgroup$
– pendermath
Jan 5 at 19:37
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38
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.
$endgroup$
– Matt Samuel
Jan 5 at 19:38
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39
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?”
$endgroup$
– Matt Samuel
Jan 5 at 19:39
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39
I am not assuming that is a subset of a group. Maybe I missed the question.
$endgroup$
– pendermath
Jan 5 at 19:39

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 $G$.
$endgroup$
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 $G$.
$endgroup$
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 $G$.
$endgroup$
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 $G$.
add a comment 
add a comment 
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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.
$endgroup$
– Axion004
Jan 5 at 14:29