Mixing
Evaluate $a_1cdot dots cdot a_n$ where $G={a_1,…,a_n}$ is an abelian group with no element $ane e$ such...
Let $G={a_1,...,a_n}$ be a finite abelian group such that $nexists ane e$ with $a^2=e$
Evaluate $a_1cdotdotscdot a_n$
I thaught that a finite abelian group with the property of $a^2=eimplies a=e$ is isomorphic to $Bbb Z/pBbb Z$ with $p$ a prime, but that's not true cause $Bbb Z/9Bbb Z$ verifies this property and $9$ is not prime....
Anyway, how can one solve this exercise without using an isomorphism with a subgroup of $Bbb Z$?
group-theory finite-groups abelian-groups
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
add a comment |
Let $G={a_1,...,a_n}$ be a finite abelian group such that $nexists ane e$ with $a^2=e$
Evaluate $a_1cdotdotscdot a_n$
I thaught that a finite abelian group with the property of $a^2=eimplies a=e$ is isomorphic to $Bbb Z/pBbb Z$ with $p$ a prime, but that's not true cause $Bbb Z/9Bbb Z$ verifies this property and $9$ is not prime....
Anyway, how can one solve this exercise without using an isomorphism with a subgroup of $Bbb Z$?
group-theory finite-groups abelian-groups
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
add a comment |
Let $G={a_1,...,a_n}$ be a finite abelian group such that $nexists ane e$ with $a^2=e$
Evaluate $a_1cdotdotscdot a_n$
I thaught that a finite abelian group with the property of $a^2=eimplies a=e$ is isomorphic to $Bbb Z/pBbb Z$ with $p$ a prime, but that's not true cause $Bbb Z/9Bbb Z$ verifies this property and $9$ is not prime....
Anyway, how can one solve this exercise without using an isomorphism with a subgroup of $Bbb Z$?
group-theory finite-groups abelian-groups
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
Let $G={a_1,...,a_n}$ be a finite abelian group such that $nexists ane e$ with $a^2=e$
Evaluate $a_1cdotdotscdot a_n$
I thaught that a finite abelian group with the property of $a^2=eimplies a=e$ is isomorphic to $Bbb Z/pBbb Z$ with $p$ a prime, but that's not true cause $Bbb Z/9Bbb Z$ verifies this property and $9$ is not prime....
Anyway, how can one solve this exercise without using an isomorphism with a subgroup of $Bbb Z$?
group-theory finite-groups abelian-groups
group-theory finite-groups abelian-groups
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
edited Nov 20 '18 at 17:37
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
asked Nov 20 '18 at 17:34
John Cataldo
1,0061216
1,0061216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint:
In the product $;a_1cdotldotscdot a_n;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)
Finally, unrelated to the question itself but also interesting: can you characterize the number $;n;$ ?
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
add a comment |
Let $x=a_1...a_n$
Claim: $n$ is odd (and the number of non identity elements is even)
If $xne e$ then $x^2ne eimplies (a_1...a_n)(a_1...a_n)ne e$
But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $ene a_1a_1^{-1}cdot...cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}cdot e=e$
which is absurd. So $x=e$
Now we need to prove the claim:
Consider the partition of $G$ by the classes of equivalence $xsim yiff x=y text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class ${a=a^{-1}}$ then $a^2=eimplies a=e$. So $n$ is odd.
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
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
Hint:
In the product $;a_1cdotldotscdot a_n;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)
Finally, unrelated to the question itself but also interesting: can you characterize the number $;n;$ ?
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
add a comment |
Hint:
In the product $;a_1cdotldotscdot a_n;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)
Finally, unrelated to the question itself but also interesting: can you characterize the number $;n;$ ?
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
add a comment |
Hint:
In the product $;a_1cdotldotscdot a_n;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)
Finally, unrelated to the question itself but also interesting: can you characterize the number $;n;$ ?
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
Hint:
In the product $;a_1cdotldotscdot a_n;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)
Finally, unrelated to the question itself but also interesting: can you characterize the number $;n;$ ?
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
answered Nov 20 '18 at 17:37
DonAntonio
177k1491225
177k1491225
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
add a comment |
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
Yes I did that already, should have included it in the question. But I never use the property $a^2ne e forall ane e$
– John Cataldo
Nov 20 '18 at 17:38
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
I cannot characterize $n$ if you mean to infer that $n$ is prime, cause it doesn't hold for $Bbb Z/9Bbb Z$
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
Oh $n$ should be odd, is that it?
– John Cataldo
Nov 20 '18 at 17:41
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
@JohnCataldo Yes, that is it...and you have to have used $;a^2neq e;$, otherwise you can't pair up each element of the group with other element ...and this is important also for the above characterization of $;n;$ , certainly.
– DonAntonio
Nov 20 '18 at 17:51
add a comment |
Let $x=a_1...a_n$
Claim: $n$ is odd (and the number of non identity elements is even)
If $xne e$ then $x^2ne eimplies (a_1...a_n)(a_1...a_n)ne e$
But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $ene a_1a_1^{-1}cdot...cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}cdot e=e$
which is absurd. So $x=e$
Now we need to prove the claim:
Consider the partition of $G$ by the classes of equivalence $xsim yiff x=y text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class ${a=a^{-1}}$ then $a^2=eimplies a=e$. So $n$ is odd.
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
add a comment |
Let $x=a_1...a_n$
Claim: $n$ is odd (and the number of non identity elements is even)
If $xne e$ then $x^2ne eimplies (a_1...a_n)(a_1...a_n)ne e$
But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $ene a_1a_1^{-1}cdot...cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}cdot e=e$
which is absurd. So $x=e$
Now we need to prove the claim:
Consider the partition of $G$ by the classes of equivalence $xsim yiff x=y text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class ${a=a^{-1}}$ then $a^2=eimplies a=e$. So $n$ is odd.
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
add a comment |
Let $x=a_1...a_n$
Claim: $n$ is odd (and the number of non identity elements is even)
If $xne e$ then $x^2ne eimplies (a_1...a_n)(a_1...a_n)ne e$
But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $ene a_1a_1^{-1}cdot...cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}cdot e=e$
which is absurd. So $x=e$
Now we need to prove the claim:
Consider the partition of $G$ by the classes of equivalence $xsim yiff x=y text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class ${a=a^{-1}}$ then $a^2=eimplies a=e$. So $n$ is odd.
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
Let $x=a_1...a_n$
Claim: $n$ is odd (and the number of non identity elements is even)
If $xne e$ then $x^2ne eimplies (a_1...a_n)(a_1...a_n)ne e$
But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $ene a_1a_1^{-1}cdot...cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}cdot e=e$
which is absurd. So $x=e$
Now we need to prove the claim:
Consider the partition of $G$ by the classes of equivalence $xsim yiff x=y text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class ${a=a^{-1}}$ then $a^2=eimplies a=e$. So $n$ is odd.
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
edited Nov 22 '18 at 12:36
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
answered Nov 20 '18 at 18:00
John Cataldo
1,0061216
1,0061216
add a comment |
add a comment |
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