Mixing
If $G$ is a group and $gin G$, then for all $n,minmathbb{Z}$, we have the following properties.
$begingroup$
Proof attempt:
(a) $g^ng^m=g^{n+m}$
(b) $(g^n)^{-1}=g^{-n}$
Proof(informal rough draft).
(a) Since $gin G$, we can rewrite $g^n=gg...g$ for n-factors of $g$ and $g^m=gg...g$ for m-factors of $g$, since rules of exponents still hold for groups(given definition in the book). If $G$ is under a group under addition, we can add the exponents to obtain $g^ng^m=g^{n+m}$.
Have not yet began (b).
Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.
group-theory proof-verification
asked Feb 1 at 2:49
RyanRyan
1898
$endgroup$
|
show 2 more comments
$begingroup$
Proof attempt:
(a) $g^ng^m=g^{n+m}$
(b) $(g^n)^{-1}=g^{-n}$
Proof(informal rough draft).
(a) Since $gin G$, we can rewrite $g^n=gg...g$ for n-factors of $g$ and $g^m=gg...g$ for m-factors of $g$, since rules of exponents still hold for groups(given definition in the book). If $G$ is under a group under addition, we can add the exponents to obtain $g^ng^m=g^{n+m}$.
Have not yet began (b).
Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.
group-theory proof-verification
asked Feb 1 at 2:49
RyanRyan
1898
$endgroup$
$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
1
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
1
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05
|
show 2 more comments
$begingroup$
Proof attempt:
(a) $g^ng^m=g^{n+m}$
(b) $(g^n)^{-1}=g^{-n}$
Proof(informal rough draft).
(a) Since $gin G$, we can rewrite $g^n=gg...g$ for n-factors of $g$ and $g^m=gg...g$ for m-factors of $g$, since rules of exponents still hold for groups(given definition in the book). If $G$ is under a group under addition, we can add the exponents to obtain $g^ng^m=g^{n+m}$.
Have not yet began (b).
Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.
group-theory proof-verification
asked Feb 1 at 2:49
RyanRyan
1898
$endgroup$
Proof attempt:
(a) $g^ng^m=g^{n+m}$
(b) $(g^n)^{-1}=g^{-n}$
Proof(informal rough draft).
(a) Since $gin G$, we can rewrite $g^n=gg...g$ for n-factors of $g$ and $g^m=gg...g$ for m-factors of $g$, since rules of exponents still hold for groups(given definition in the book). If $G$ is under a group under addition, we can add the exponents to obtain $g^ng^m=g^{n+m}$.
Have not yet began (b).
Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.
group-theory proof-verification
group-theory proof-verification
asked Feb 1 at 2:49
RyanRyan
1898
asked Feb 1 at 2:49
RyanRyan
1898
asked Feb 1 at 2:49
RyanRyan
1898
asked Feb 1 at 2:49
RyanRyan
1898
asked Feb 1 at 2:49
RyanRyan
1898
1898
$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
1
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
1
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05
|
show 2 more comments
$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
1
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
1
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05
$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
1
1
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
1
1
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05
|
show 2 more comments
1 Answer
1
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oldest
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$begingroup$
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
$endgroup$
add a comment |
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$begingroup$
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
$endgroup$
add a comment |
$begingroup$
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
$endgroup$
add a comment |
$begingroup$
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
$endgroup$
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
answered Feb 1 at 12:11
Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda
1164
1164
add a comment |
add a comment |
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$begingroup$
If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument.
$endgroup$
– T. Bongers
Feb 1 at 2:54
$begingroup$
You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$.
$endgroup$
– J. W. Tanner
Feb 1 at 2:57
$begingroup$
How does your book define $g^n$ for $n in mathbb{Z}$?
$endgroup$
– Theo Bendit
Feb 1 at 2:57
1
$begingroup$
For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element.
$endgroup$
– angryavian
Feb 1 at 3:01
1
$begingroup$
@Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$?
$endgroup$
– Theo Bendit
Feb 1 at 3:05