Mixing
Understanding a Cyclic Group Proof
Clash Royale CLAN TAG#URR8PPP
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For one of my questions, the following answer was given by Suzet:
The thing is: I do not see why... Is it because of inverses or because having a negative could make $i-j$ positive? I asked this in the comment but no one replied yet. Does anyone know? :) I feel like I am missing something very trivial.
group-theory proof-explanation cyclic-groups
asked Aug 12 at 23:15
numericalorange
1,286110
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up vote
2
down vote
favorite
For one of my questions, the following answer was given by Suzet:
The thing is: I do not see why... Is it because of inverses or because having a negative could make $i-j$ positive? I asked this in the comment but no one replied yet. Does anyone know? :) I feel like I am missing something very trivial.
group-theory proof-explanation cyclic-groups
asked Aug 12 at 23:15
numericalorange
1,286110
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For one of my questions, the following answer was given by Suzet:
The thing is: I do not see why... Is it because of inverses or because having a negative could make $i-j$ positive? I asked this in the comment but no one replied yet. Does anyone know? :) I feel like I am missing something very trivial.
group-theory proof-explanation cyclic-groups
asked Aug 12 at 23:15
numericalorange
1,286110
For one of my questions, the following answer was given by Suzet:
The thing is: I do not see why... Is it because of inverses or because having a negative could make $i-j$ positive? I asked this in the comment but no one replied yet. Does anyone know? :) I feel like I am missing something very trivial.
group-theory proof-explanation cyclic-groups
asked Aug 12 at 23:15
numericalorange
1,286110
asked Aug 12 at 23:15
numericalorange
1,286110
asked Aug 12 at 23:15
numericalorange
1,286110
asked Aug 12 at 23:15
numericalorange
1,286110
1,286110
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
4
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accepted
A variant: $enspace$if $a^-n=e;(ninmathbf N)$, then $; e=e^-1=(a^-n)^-1=a^(-n)(-1)=a^n.$
answered Aug 12 at 23:58
Bernard
111k635103
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
add a comment |Â
up vote
3
down vote
Suppose that for a positive integer $n$, we have $a^-n = e$. Then multiplying by $a$ on each side, we get $a = a^-n+1$. Then we simply have:
$$
a^n = (a^-n+1)^n = a^-n^2+n = a^-n(n-1) = (a^-n)^n-1 = e^n-1 = e
$$
answered Aug 12 at 23:22
Sambo
1,3631427
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
A variant: $enspace$if $a^-n=e;(ninmathbf N)$, then $; e=e^-1=(a^-n)^-1=a^(-n)(-1)=a^n.$
answered Aug 12 at 23:58
Bernard
111k635103
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
add a comment |Â
up vote
4
down vote
accepted
A variant: $enspace$if $a^-n=e;(ninmathbf N)$, then $; e=e^-1=(a^-n)^-1=a^(-n)(-1)=a^n.$
answered Aug 12 at 23:58
Bernard
111k635103
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
A variant: $enspace$if $a^-n=e;(ninmathbf N)$, then $; e=e^-1=(a^-n)^-1=a^(-n)(-1)=a^n.$
answered Aug 12 at 23:58
Bernard
111k635103
A variant: $enspace$if $a^-n=e;(ninmathbf N)$, then $; e=e^-1=(a^-n)^-1=a^(-n)(-1)=a^n.$
answered Aug 12 at 23:58
Bernard
111k635103
answered Aug 12 at 23:58
Bernard
111k635103
answered Aug 12 at 23:58
Bernard
111k635103
answered Aug 12 at 23:58
Bernard
111k635103
111k635103
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
add a comment |Â
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
Very simple answer and super easy to understand! Thank you very much!!
– numericalorange
Aug 12 at 23:59
add a comment |Â
up vote
3
down vote
Suppose that for a positive integer $n$, we have $a^-n = e$. Then multiplying by $a$ on each side, we get $a = a^-n+1$. Then we simply have:
$$
a^n = (a^-n+1)^n = a^-n^2+n = a^-n(n-1) = (a^-n)^n-1 = e^n-1 = e
$$
answered Aug 12 at 23:22
Sambo
1,3631427
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
add a comment |Â
up vote
3
down vote
Suppose that for a positive integer $n$, we have $a^-n = e$. Then multiplying by $a$ on each side, we get $a = a^-n+1$. Then we simply have:
$$
a^n = (a^-n+1)^n = a^-n^2+n = a^-n(n-1) = (a^-n)^n-1 = e^n-1 = e
$$
answered Aug 12 at 23:22
Sambo
1,3631427
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Suppose that for a positive integer $n$, we have $a^-n = e$. Then multiplying by $a$ on each side, we get $a = a^-n+1$. Then we simply have:
$$
a^n = (a^-n+1)^n = a^-n^2+n = a^-n(n-1) = (a^-n)^n-1 = e^n-1 = e
$$
answered Aug 12 at 23:22
Sambo
1,3631427
Suppose that for a positive integer $n$, we have $a^-n = e$. Then multiplying by $a$ on each side, we get $a = a^-n+1$. Then we simply have:
$$
a^n = (a^-n+1)^n = a^-n^2+n = a^-n(n-1) = (a^-n)^n-1 = e^n-1 = e
$$
answered Aug 12 at 23:22
Sambo
1,3631427
edited Aug 13 at 0:22
answered Aug 12 at 23:22
Sambo
1,3631427
answered Aug 12 at 23:22
Sambo
1,3631427
answered Aug 12 at 23:22
Sambo
1,3631427
1,3631427
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
add a comment |Â
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
1
1
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Wouldn't it be simpler to just take inverses on both sides of $a^-n=e$? Also, you are assuming $a^-n=e$, but in your third equality you seem to be using $a^n=e$, which needs justification, as that is supposed to be what you are trying to establish.
– Arturo Magidin
Aug 12 at 23:55
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
Thank you both for replying and helping me the best way you can. I was also confused why $a^n=e$ was used. So can I just say that if $n$ is a positive integer such that $a^-n=e$, then by inverses, $a^na^-n=a^ne$ or $e=a^n$ which is a contradiction?
– numericalorange
Aug 12 at 23:58
1
1
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
@numericalorange: Are you addressing me, or Sambo? Multiplying both sides of $a^-n=e$ by $a^n$ also gives that if $a^-n=e$, then $a^n=e$, yes.
– Arturo Magidin
Aug 13 at 0:14
1
1
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
@ArturoMagidin You're absolutely right; I messed up. I'll edit it now. Bernard's answer above is indeed a much quicker solution.
– Sambo
Aug 13 at 0:21
add a comment |Â
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