Mixing
Subgroups. Prove Theorem: If $G$ is a group, then $Gleq G$.
$begingroup$
Theorem: If $G$ is a group, prove $Gleq G$.
This proof seems straight forward, but I am having a hard time forming a coherent proof.
Ideas: We assume $G$ is a group. Then we prove any element in $G$ satisfies the following properties of a group. Let $xin G$. Show that $x$ satisfies the binary operation and has an identity element in $G$?
I am brand new to proofs.
abstract-algebra
asked Feb 1 at 10:31
RyanRyan
1898
$endgroup$
|
show 1 more comment
$begingroup$
Theorem: If $G$ is a group, prove $Gleq G$.
This proof seems straight forward, but I am having a hard time forming a coherent proof.
Ideas: We assume $G$ is a group. Then we prove any element in $G$ satisfies the following properties of a group. Let $xin G$. Show that $x$ satisfies the binary operation and has an identity element in $G$?
I am brand new to proofs.
abstract-algebra
asked Feb 1 at 10:31
RyanRyan
1898
$endgroup$
5
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
1
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
4
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44
|
show 1 more comment
$begingroup$
Theorem: If $G$ is a group, prove $Gleq G$.
This proof seems straight forward, but I am having a hard time forming a coherent proof.
Ideas: We assume $G$ is a group. Then we prove any element in $G$ satisfies the following properties of a group. Let $xin G$. Show that $x$ satisfies the binary operation and has an identity element in $G$?
I am brand new to proofs.
abstract-algebra
asked Feb 1 at 10:31
RyanRyan
1898
$endgroup$
Theorem: If $G$ is a group, prove $Gleq G$.
This proof seems straight forward, but I am having a hard time forming a coherent proof.
Ideas: We assume $G$ is a group. Then we prove any element in $G$ satisfies the following properties of a group. Let $xin G$. Show that $x$ satisfies the binary operation and has an identity element in $G$?
I am brand new to proofs.
abstract-algebra
abstract-algebra
asked Feb 1 at 10:31
RyanRyan
1898
asked Feb 1 at 10:31
RyanRyan
1898
asked Feb 1 at 10:31
RyanRyan
1898
asked Feb 1 at 10:31
RyanRyan
1898
asked Feb 1 at 10:31
RyanRyan
1898
1898
5
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
1
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
4
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44
|
show 1 more comment
5
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
1
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
4
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44
5
5
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
1
1
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
4
4
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Showing $G leq G$ is as trivial as showing that $G$ is a group, when you already know that $G$ is a group.
For a set $H subseteq G$ to be a subgroup of $G$ it must satisfy the
following criteria:
$bullet space$ Closure: $forall h_1,h_2 in H: space$ $h_1
cdot h_2 in H$, where $cdot$ is $G$'s binary operation.
$bullet space$ Identity: $space exists e in H$
$bullet space$ Inverses:$space forall hin H, exists h^{-1}
in H$ s.t. $h cdot h^{-1}=e.$
In your case, when $H=G subseteq G$, all criteria are satisfied by definition:
$bullet space$Closure: $forall g_1,g_2 in G$: $space g_1 cdot g_2 in G$, since $G$ is a group.
$bullet space$ Identinty: $e in G$, since $G$ is a group.
$bullet space$ Inverses: $forall g in G, exists g^{-1}$ s.t. $g cdot g^{-1}=e$, since $G$ is a group.
To summarize, for a group to be a subgroup of itself, the only necessary and sufficient condition is to be a group. That's why the whole group is called the trivial subgroup of itself.
answered Feb 1 at 13:42
JevautJevaut
1,168312
$endgroup$
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
add a comment |
$begingroup$
Here is my suggestion for a "standard solution", discussed in the comments.
Claim: Let $G$ be a group. Then $G$ is a subgroup of $G$.
Proof: We use the following result, which is usually available in the lecture right after the definition of a subgroup: a non-empty subset $Ssubseteq G$ is a subgroup, if and only if $xyin S$ for all $x,yin S$ and $x^{-1}in S$ for all $xin S$. Both conditions are satisfied for $S=G$ by the very definition of a group, i.e., because $G$ is a group.
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Showing $G leq G$ is as trivial as showing that $G$ is a group, when you already know that $G$ is a group.
For a set $H subseteq G$ to be a subgroup of $G$ it must satisfy the
following criteria:
$bullet space$ Closure: $forall h_1,h_2 in H: space$ $h_1
cdot h_2 in H$, where $cdot$ is $G$'s binary operation.
$bullet space$ Identity: $space exists e in H$
$bullet space$ Inverses:$space forall hin H, exists h^{-1}
in H$ s.t. $h cdot h^{-1}=e.$
In your case, when $H=G subseteq G$, all criteria are satisfied by definition:
$bullet space$Closure: $forall g_1,g_2 in G$: $space g_1 cdot g_2 in G$, since $G$ is a group.
$bullet space$ Identinty: $e in G$, since $G$ is a group.
$bullet space$ Inverses: $forall g in G, exists g^{-1}$ s.t. $g cdot g^{-1}=e$, since $G$ is a group.
To summarize, for a group to be a subgroup of itself, the only necessary and sufficient condition is to be a group. That's why the whole group is called the trivial subgroup of itself.
answered Feb 1 at 13:42
JevautJevaut
1,168312
$endgroup$
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
add a comment |
$begingroup$
Showing $G leq G$ is as trivial as showing that $G$ is a group, when you already know that $G$ is a group.
For a set $H subseteq G$ to be a subgroup of $G$ it must satisfy the
following criteria:
$bullet space$ Closure: $forall h_1,h_2 in H: space$ $h_1
cdot h_2 in H$, where $cdot$ is $G$'s binary operation.
$bullet space$ Identity: $space exists e in H$
$bullet space$ Inverses:$space forall hin H, exists h^{-1}
in H$ s.t. $h cdot h^{-1}=e.$
In your case, when $H=G subseteq G$, all criteria are satisfied by definition:
$bullet space$Closure: $forall g_1,g_2 in G$: $space g_1 cdot g_2 in G$, since $G$ is a group.
$bullet space$ Identinty: $e in G$, since $G$ is a group.
$bullet space$ Inverses: $forall g in G, exists g^{-1}$ s.t. $g cdot g^{-1}=e$, since $G$ is a group.
To summarize, for a group to be a subgroup of itself, the only necessary and sufficient condition is to be a group. That's why the whole group is called the trivial subgroup of itself.
answered Feb 1 at 13:42
JevautJevaut
1,168312
$endgroup$
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
add a comment |
$begingroup$
Showing $G leq G$ is as trivial as showing that $G$ is a group, when you already know that $G$ is a group.
For a set $H subseteq G$ to be a subgroup of $G$ it must satisfy the
following criteria:
$bullet space$ Closure: $forall h_1,h_2 in H: space$ $h_1
cdot h_2 in H$, where $cdot$ is $G$'s binary operation.
$bullet space$ Identity: $space exists e in H$
$bullet space$ Inverses:$space forall hin H, exists h^{-1}
in H$ s.t. $h cdot h^{-1}=e.$
In your case, when $H=G subseteq G$, all criteria are satisfied by definition:
$bullet space$Closure: $forall g_1,g_2 in G$: $space g_1 cdot g_2 in G$, since $G$ is a group.
$bullet space$ Identinty: $e in G$, since $G$ is a group.
$bullet space$ Inverses: $forall g in G, exists g^{-1}$ s.t. $g cdot g^{-1}=e$, since $G$ is a group.
To summarize, for a group to be a subgroup of itself, the only necessary and sufficient condition is to be a group. That's why the whole group is called the trivial subgroup of itself.
answered Feb 1 at 13:42
JevautJevaut
1,168312
$endgroup$
Showing $G leq G$ is as trivial as showing that $G$ is a group, when you already know that $G$ is a group.
For a set $H subseteq G$ to be a subgroup of $G$ it must satisfy the
following criteria:
$bullet space$ Closure: $forall h_1,h_2 in H: space$ $h_1
cdot h_2 in H$, where $cdot$ is $G$'s binary operation.
$bullet space$ Identity: $space exists e in H$
$bullet space$ Inverses:$space forall hin H, exists h^{-1}
in H$ s.t. $h cdot h^{-1}=e.$
In your case, when $H=G subseteq G$, all criteria are satisfied by definition:
$bullet space$Closure: $forall g_1,g_2 in G$: $space g_1 cdot g_2 in G$, since $G$ is a group.
$bullet space$ Identinty: $e in G$, since $G$ is a group.
$bullet space$ Inverses: $forall g in G, exists g^{-1}$ s.t. $g cdot g^{-1}=e$, since $G$ is a group.
To summarize, for a group to be a subgroup of itself, the only necessary and sufficient condition is to be a group. That's why the whole group is called the trivial subgroup of itself.
answered Feb 1 at 13:42
JevautJevaut
1,168312
answered Feb 1 at 13:42
JevautJevaut
1,168312
answered Feb 1 at 13:42
JevautJevaut
1,168312
answered Feb 1 at 13:42
JevautJevaut
1,168312
1,168312
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
add a comment |
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
Thank you for the thorough explanation and easy to follow steps.
$endgroup$
– Ryan
Feb 1 at 14:29
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
$begingroup$
You're welcome, thanks for supporting my answer.
$endgroup$
– Jevaut
Feb 1 at 15:15
add a comment |
$begingroup$
Here is my suggestion for a "standard solution", discussed in the comments.
Claim: Let $G$ be a group. Then $G$ is a subgroup of $G$.
Proof: We use the following result, which is usually available in the lecture right after the definition of a subgroup: a non-empty subset $Ssubseteq G$ is a subgroup, if and only if $xyin S$ for all $x,yin S$ and $x^{-1}in S$ for all $xin S$. Both conditions are satisfied for $S=G$ by the very definition of a group, i.e., because $G$ is a group.
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
$endgroup$
add a comment |
$begingroup$
Here is my suggestion for a "standard solution", discussed in the comments.
Claim: Let $G$ be a group. Then $G$ is a subgroup of $G$.
Proof: We use the following result, which is usually available in the lecture right after the definition of a subgroup: a non-empty subset $Ssubseteq G$ is a subgroup, if and only if $xyin S$ for all $x,yin S$ and $x^{-1}in S$ for all $xin S$. Both conditions are satisfied for $S=G$ by the very definition of a group, i.e., because $G$ is a group.
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
$endgroup$
add a comment |
$begingroup$
Here is my suggestion for a "standard solution", discussed in the comments.
Claim: Let $G$ be a group. Then $G$ is a subgroup of $G$.
Proof: We use the following result, which is usually available in the lecture right after the definition of a subgroup: a non-empty subset $Ssubseteq G$ is a subgroup, if and only if $xyin S$ for all $x,yin S$ and $x^{-1}in S$ for all $xin S$. Both conditions are satisfied for $S=G$ by the very definition of a group, i.e., because $G$ is a group.
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
$endgroup$
Here is my suggestion for a "standard solution", discussed in the comments.
Claim: Let $G$ be a group. Then $G$ is a subgroup of $G$.
Proof: We use the following result, which is usually available in the lecture right after the definition of a subgroup: a non-empty subset $Ssubseteq G$ is a subgroup, if and only if $xyin S$ for all $x,yin S$ and $x^{-1}in S$ for all $xin S$. Both conditions are satisfied for $S=G$ by the very definition of a group, i.e., because $G$ is a group.
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
edited Feb 1 at 12:37
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
answered Feb 1 at 10:48
Dietrich BurdeDietrich Burde
81.9k649107
81.9k649107
add a comment |
add a comment |
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5
$begingroup$
$G$ is a group. $G$ is a subset of $G$. That's pretty much all there is.
$endgroup$
– freakish
Feb 1 at 10:34
1
$begingroup$
Lol, thank you.
$endgroup$
– Ryan
Feb 1 at 10:34
4
$begingroup$
Here it seems important that you find out why you had a hard time. So the claim is: if $G$ is a group, then $G$ is a subgroup of itself. Now can you formulate the axioms (or criteria) for being a subgroup? Then the next step is to verify them just by using the axioms from the assumption.
$endgroup$
– Dietrich Burde
Feb 1 at 10:37
$begingroup$
@DietrichBurde Yes, I agree. I wrote down the intuitive answer, but just wanted more so to confirm I wasn't being "lazy." Lack of sleep leads to non-thinking or overthinking. Thank you for your response.
$endgroup$
– Ryan
Feb 1 at 10:42
$begingroup$
If you want I can write down a "formal standard solution". Of course, you know it already.
$endgroup$
– Dietrich Burde
Feb 1 at 10:44