Mixing
If + is associative, prove that +' is associative
$begingroup$
I'm doing some self-study of elliptic curves and I'm having some difficulty with a problem. Here is the statement:
Let $S$ be a set with a composition law $ast$ which has the following two properties
i) $P ast Q = Q ast P$ for all $P,Q in S$
ii) $P ast (P ast Q) = Q$ for all $P,Q in S$.
We define $+'$ to be $P +' Q = mathcal{O}' ast (P ast Q)$ where $mathcal{O}' in S$. Assume that $+$ is associative. Prove that $+'$ is associative.
I have earlier proven a result that says that if $+$ is associative then
$R ast (mathcal{O} ast (P ast Q)) = P ast (mathcal{O} ast (Q ast R))$ for all $P,Q,R in S$ and $mathcal{O} in S$ is fixed.
Note that $+$ is defined similarly by $P + Q = mathcal{O} ast (P ast Q)$.
I know I need to stem my proof from this previous result but picking a specific $P,Q,R in S$ which cancels out $mathcal{O}$ and replaces it with $mathcal{O}'$ but I'm lost as to how to do this.
abstract-algebra
edited Feb 1 at 5:28
Brandon Carter
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
$endgroup$
add a comment |
$begingroup$
I'm doing some self-study of elliptic curves and I'm having some difficulty with a problem. Here is the statement:
Let $S$ be a set with a composition law $ast$ which has the following two properties
i) $P ast Q = Q ast P$ for all $P,Q in S$
ii) $P ast (P ast Q) = Q$ for all $P,Q in S$.
We define $+'$ to be $P +' Q = mathcal{O}' ast (P ast Q)$ where $mathcal{O}' in S$. Assume that $+$ is associative. Prove that $+'$ is associative.
I have earlier proven a result that says that if $+$ is associative then
$R ast (mathcal{O} ast (P ast Q)) = P ast (mathcal{O} ast (Q ast R))$ for all $P,Q,R in S$ and $mathcal{O} in S$ is fixed.
Note that $+$ is defined similarly by $P + Q = mathcal{O} ast (P ast Q)$.
I know I need to stem my proof from this previous result but picking a specific $P,Q,R in S$ which cancels out $mathcal{O}$ and replaces it with $mathcal{O}'$ but I'm lost as to how to do this.
abstract-algebra
edited Feb 1 at 5:28
Brandon Carter
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
$endgroup$
1
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06
add a comment |
$begingroup$
I'm doing some self-study of elliptic curves and I'm having some difficulty with a problem. Here is the statement:
Let $S$ be a set with a composition law $ast$ which has the following two properties
i) $P ast Q = Q ast P$ for all $P,Q in S$
ii) $P ast (P ast Q) = Q$ for all $P,Q in S$.
We define $+'$ to be $P +' Q = mathcal{O}' ast (P ast Q)$ where $mathcal{O}' in S$. Assume that $+$ is associative. Prove that $+'$ is associative.
I have earlier proven a result that says that if $+$ is associative then
$R ast (mathcal{O} ast (P ast Q)) = P ast (mathcal{O} ast (Q ast R))$ for all $P,Q,R in S$ and $mathcal{O} in S$ is fixed.
Note that $+$ is defined similarly by $P + Q = mathcal{O} ast (P ast Q)$.
I know I need to stem my proof from this previous result but picking a specific $P,Q,R in S$ which cancels out $mathcal{O}$ and replaces it with $mathcal{O}'$ but I'm lost as to how to do this.
abstract-algebra
edited Feb 1 at 5:28
Brandon Carter
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
$endgroup$
I'm doing some self-study of elliptic curves and I'm having some difficulty with a problem. Here is the statement:
Let $S$ be a set with a composition law $ast$ which has the following two properties
i) $P ast Q = Q ast P$ for all $P,Q in S$
ii) $P ast (P ast Q) = Q$ for all $P,Q in S$.
We define $+'$ to be $P +' Q = mathcal{O}' ast (P ast Q)$ where $mathcal{O}' in S$. Assume that $+$ is associative. Prove that $+'$ is associative.
I have earlier proven a result that says that if $+$ is associative then
$R ast (mathcal{O} ast (P ast Q)) = P ast (mathcal{O} ast (Q ast R))$ for all $P,Q,R in S$ and $mathcal{O} in S$ is fixed.
Note that $+$ is defined similarly by $P + Q = mathcal{O} ast (P ast Q)$.
I know I need to stem my proof from this previous result but picking a specific $P,Q,R in S$ which cancels out $mathcal{O}$ and replaces it with $mathcal{O}'$ but I'm lost as to how to do this.
abstract-algebra
abstract-algebra
edited Feb 1 at 5:28
Brandon Carter
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
edited Feb 1 at 5:28
Brandon Carter
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
edited Feb 1 at 5:28
Brandon Carter
7,31822538
edited Feb 1 at 5:28
Brandon Carter
7,31822538
edited Feb 1 at 5:28
Brandon Carter
7,31822538
7,31822538
asked Feb 1 at 3:02
enigmaenigma
353
asked Feb 1 at 3:02
enigmaenigma
353
asked Feb 1 at 3:02
enigmaenigma
353
353
1
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06
add a comment |
1
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06
1
1
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Okay here goes nothing:
If we assume $+$ to be associative, then we have $R*(O*(Q*P)) = P*(O*(Q*R))$. So means we can swap elements provided there is an $O$ in the correct position.
On the LHS, $R*(O*(Q*P)) = R*(O*(O*(O*(P*Q))) = O*(O*(O*(R*(P*Q)))) = O*(R*(P*Q))$
Here I used (ii) and the associativity of $O$.
Then multiply by $O$ to get $(R*(P*Q))$. Then multiply by $O'$ to get $O'*(R*(P*Q))$. Effectively replacing $O$ by $O'$.
Repeat the steps above to swap $R$ and $O'$ again: $O'*(R*(P*Q)) = O'*(O*(O*(R*(P*Q)))) = R*(O*(O*(O'*(P*Q))) = R*(O'*(Q*P))$
We can do the same steps on the RHS while preserving inequality.
So we get $R*(O'*(Q*P)) = P*(O'*(Q*R))$
answered Feb 5 at 6:14
eatfoodeatfood
3318
$endgroup$
add a comment |
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$begingroup$
Okay here goes nothing:
If we assume $+$ to be associative, then we have $R*(O*(Q*P)) = P*(O*(Q*R))$. So means we can swap elements provided there is an $O$ in the correct position.
On the LHS, $R*(O*(Q*P)) = R*(O*(O*(O*(P*Q))) = O*(O*(O*(R*(P*Q)))) = O*(R*(P*Q))$
Here I used (ii) and the associativity of $O$.
Then multiply by $O$ to get $(R*(P*Q))$. Then multiply by $O'$ to get $O'*(R*(P*Q))$. Effectively replacing $O$ by $O'$.
Repeat the steps above to swap $R$ and $O'$ again: $O'*(R*(P*Q)) = O'*(O*(O*(R*(P*Q)))) = R*(O*(O*(O'*(P*Q))) = R*(O'*(Q*P))$
We can do the same steps on the RHS while preserving inequality.
So we get $R*(O'*(Q*P)) = P*(O'*(Q*R))$
answered Feb 5 at 6:14
eatfoodeatfood
3318
$endgroup$
add a comment |
$begingroup$
Okay here goes nothing:
If we assume $+$ to be associative, then we have $R*(O*(Q*P)) = P*(O*(Q*R))$. So means we can swap elements provided there is an $O$ in the correct position.
On the LHS, $R*(O*(Q*P)) = R*(O*(O*(O*(P*Q))) = O*(O*(O*(R*(P*Q)))) = O*(R*(P*Q))$
Here I used (ii) and the associativity of $O$.
Then multiply by $O$ to get $(R*(P*Q))$. Then multiply by $O'$ to get $O'*(R*(P*Q))$. Effectively replacing $O$ by $O'$.
Repeat the steps above to swap $R$ and $O'$ again: $O'*(R*(P*Q)) = O'*(O*(O*(R*(P*Q)))) = R*(O*(O*(O'*(P*Q))) = R*(O'*(Q*P))$
We can do the same steps on the RHS while preserving inequality.
So we get $R*(O'*(Q*P)) = P*(O'*(Q*R))$
answered Feb 5 at 6:14
eatfoodeatfood
3318
$endgroup$
add a comment |
$begingroup$
Okay here goes nothing:
If we assume $+$ to be associative, then we have $R*(O*(Q*P)) = P*(O*(Q*R))$. So means we can swap elements provided there is an $O$ in the correct position.
On the LHS, $R*(O*(Q*P)) = R*(O*(O*(O*(P*Q))) = O*(O*(O*(R*(P*Q)))) = O*(R*(P*Q))$
Here I used (ii) and the associativity of $O$.
Then multiply by $O$ to get $(R*(P*Q))$. Then multiply by $O'$ to get $O'*(R*(P*Q))$. Effectively replacing $O$ by $O'$.
Repeat the steps above to swap $R$ and $O'$ again: $O'*(R*(P*Q)) = O'*(O*(O*(R*(P*Q)))) = R*(O*(O*(O'*(P*Q))) = R*(O'*(Q*P))$
We can do the same steps on the RHS while preserving inequality.
So we get $R*(O'*(Q*P)) = P*(O'*(Q*R))$
answered Feb 5 at 6:14
eatfoodeatfood
3318
$endgroup$
Okay here goes nothing:
If we assume $+$ to be associative, then we have $R*(O*(Q*P)) = P*(O*(Q*R))$. So means we can swap elements provided there is an $O$ in the correct position.
On the LHS, $R*(O*(Q*P)) = R*(O*(O*(O*(P*Q))) = O*(O*(O*(R*(P*Q)))) = O*(R*(P*Q))$
Here I used (ii) and the associativity of $O$.
Then multiply by $O$ to get $(R*(P*Q))$. Then multiply by $O'$ to get $O'*(R*(P*Q))$. Effectively replacing $O$ by $O'$.
Repeat the steps above to swap $R$ and $O'$ again: $O'*(R*(P*Q)) = O'*(O*(O*(R*(P*Q)))) = R*(O*(O*(O'*(P*Q))) = R*(O'*(Q*P))$
We can do the same steps on the RHS while preserving inequality.
So we get $R*(O'*(Q*P)) = P*(O'*(Q*R))$
answered Feb 5 at 6:14
eatfoodeatfood
3318
answered Feb 5 at 6:14
eatfoodeatfood
3318
answered Feb 5 at 6:14
eatfoodeatfood
3318
answered Feb 5 at 6:14
eatfoodeatfood
3318
3318
add a comment |
add a comment |
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1
$begingroup$
You defined what $+'$ is, but not what $+$ is.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 3:04
$begingroup$
Thanks for the catch @LordSharktheUnknown !
$endgroup$
– enigma
Feb 1 at 3:06