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If $alpha$ and $beta$ are disjoint, so are $alpha$ and $beta^k$?
Show $pialphapi^{-1}$ and $pibetapi^{-1}$ are disjoint given $alpha$ and $beta$ are disjoint for an arbitrary permutation $pi$Cycles of odd length: $alpha^2=beta^2 implies alpha=beta$Order of a $alpha beta^{n/q}$ given the order of $alpha$ and $beta$If $alpha,betain S_n$ are disjoint, are $alpha^m, beta^n$ disjoint for any $m,ninBbb{N}$?Conditions on $alpha, beta$ under which $A rtimes_{alpha} B $ and $A rtimes_{beta} B$ are isomorphic$(alpha,beta)^{-1}=beta^{-1}alpha^{-1}$, proofdisjoint permutations commutesHow to prove that $sign(alpha beta) = sign(alpha)sign(beta)$?If $beta$ is the transposition $beta = (1, 4)$, compute both $betaalpha$ and $betaalphabeta^{−1}$ and compute orders.Commuting permutations are multiples of each other
$begingroup$
If $alpha$ and $beta$ are disjoint permutations, so are $alpha$ and $beta^k$.
I'm working on the group of permutations of a finite set $X$ with composition. The power is defined as usual using composition.
I don't know if it's true, I just found this while proving another thing. Disjoint meaning that if $alpha(x)=x$ then $beta(x)ne x$ and vice versa.
I'm trying by induction. The base case is direct. Assuming that $beta$ and $alpha^{k-1}$ are disjoint, let $iin X$. If $beta(i)ne i$ then $alpha^{k-1}(i)=i$ and $alpha(i)=i$, so aplying $alpha$ in the first equality we get $alpha^k(i) = i$.
But I'm a bit stuck in the case $beta(i)=i$. We would have $alpha^{k-1}(i)ne i$ and $alpha(i) ne i$. Assuming that $alpha^k(i) = i$ was the case then $alpha^{k-1}(alpha(i)) = i$ then $beta(alpha(i)) = alpha(i)$ but this seems to give me nothing.
Any help will be appreciated.
group-theory symmetric-groups
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
$endgroup$
add a comment |
$begingroup$
If $alpha$ and $beta$ are disjoint permutations, so are $alpha$ and $beta^k$.
I'm working on the group of permutations of a finite set $X$ with composition. The power is defined as usual using composition.
I don't know if it's true, I just found this while proving another thing. Disjoint meaning that if $alpha(x)=x$ then $beta(x)ne x$ and vice versa.
I'm trying by induction. The base case is direct. Assuming that $beta$ and $alpha^{k-1}$ are disjoint, let $iin X$. If $beta(i)ne i$ then $alpha^{k-1}(i)=i$ and $alpha(i)=i$, so aplying $alpha$ in the first equality we get $alpha^k(i) = i$.
But I'm a bit stuck in the case $beta(i)=i$. We would have $alpha^{k-1}(i)ne i$ and $alpha(i) ne i$. Assuming that $alpha^k(i) = i$ was the case then $alpha^{k-1}(alpha(i)) = i$ then $beta(alpha(i)) = alpha(i)$ but this seems to give me nothing.
Any help will be appreciated.
group-theory symmetric-groups
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
$endgroup$
add a comment |
$begingroup$
If $alpha$ and $beta$ are disjoint permutations, so are $alpha$ and $beta^k$.
I'm working on the group of permutations of a finite set $X$ with composition. The power is defined as usual using composition.
I don't know if it's true, I just found this while proving another thing. Disjoint meaning that if $alpha(x)=x$ then $beta(x)ne x$ and vice versa.
I'm trying by induction. The base case is direct. Assuming that $beta$ and $alpha^{k-1}$ are disjoint, let $iin X$. If $beta(i)ne i$ then $alpha^{k-1}(i)=i$ and $alpha(i)=i$, so aplying $alpha$ in the first equality we get $alpha^k(i) = i$.
But I'm a bit stuck in the case $beta(i)=i$. We would have $alpha^{k-1}(i)ne i$ and $alpha(i) ne i$. Assuming that $alpha^k(i) = i$ was the case then $alpha^{k-1}(alpha(i)) = i$ then $beta(alpha(i)) = alpha(i)$ but this seems to give me nothing.
Any help will be appreciated.
group-theory symmetric-groups
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
$endgroup$
If $alpha$ and $beta$ are disjoint permutations, so are $alpha$ and $beta^k$.
I'm working on the group of permutations of a finite set $X$ with composition. The power is defined as usual using composition.
I don't know if it's true, I just found this while proving another thing. Disjoint meaning that if $alpha(x)=x$ then $beta(x)ne x$ and vice versa.
I'm trying by induction. The base case is direct. Assuming that $beta$ and $alpha^{k-1}$ are disjoint, let $iin X$. If $beta(i)ne i$ then $alpha^{k-1}(i)=i$ and $alpha(i)=i$, so aplying $alpha$ in the first equality we get $alpha^k(i) = i$.
But I'm a bit stuck in the case $beta(i)=i$. We would have $alpha^{k-1}(i)ne i$ and $alpha(i) ne i$. Assuming that $alpha^k(i) = i$ was the case then $alpha^{k-1}(alpha(i)) = i$ then $beta(alpha(i)) = alpha(i)$ but this seems to give me nothing.
Any help will be appreciated.
group-theory symmetric-groups
group-theory symmetric-groups
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
asked yesterday
AnalyticHarmonyAnalyticHarmony
657313
657313
add a comment |
add a comment |
2 Answers
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$begingroup$
The statement to be proved is untrue under your definition of disjoint:
Your Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)=x$ implies $beta(x)neq x$ and vice versa.
To see this, consider $alpha=(1;2)(3)(4)$ and $beta=(1)(2)(3;4)$, $k=2$.
You get a correct result is you instead use
Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)neq x$ implies $beta(x)= x$ and vice versa.
To see this, suppose $alpha(x)neq x$. Then disjointness implies $beta(x)=x$, so $beta^k(x)=beta^{k-1}(x)=dots=beta(x)=x$. On the other hand, if $beta^k(x)neq x$, then it must be the case that $beta(x)neq x$ (since $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.
answered yesterday
Mike EarnestMike Earnest
24.3k22151
$endgroup$
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
add a comment |
$begingroup$
The set of elements that is moved by $beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $alpha$ (let's call it $A$) so if you apply $beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.
answered yesterday
gammagamma
3,39221342
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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votes
$begingroup$
The statement to be proved is untrue under your definition of disjoint:
Your Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)=x$ implies $beta(x)neq x$ and vice versa.
To see this, consider $alpha=(1;2)(3)(4)$ and $beta=(1)(2)(3;4)$, $k=2$.
You get a correct result is you instead use
Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)neq x$ implies $beta(x)= x$ and vice versa.
To see this, suppose $alpha(x)neq x$. Then disjointness implies $beta(x)=x$, so $beta^k(x)=beta^{k-1}(x)=dots=beta(x)=x$. On the other hand, if $beta^k(x)neq x$, then it must be the case that $beta(x)neq x$ (since $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.
answered yesterday
Mike EarnestMike Earnest
24.3k22151
$endgroup$
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
add a comment |
$begingroup$
The statement to be proved is untrue under your definition of disjoint:
Your Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)=x$ implies $beta(x)neq x$ and vice versa.
To see this, consider $alpha=(1;2)(3)(4)$ and $beta=(1)(2)(3;4)$, $k=2$.
You get a correct result is you instead use
Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)neq x$ implies $beta(x)= x$ and vice versa.
To see this, suppose $alpha(x)neq x$. Then disjointness implies $beta(x)=x$, so $beta^k(x)=beta^{k-1}(x)=dots=beta(x)=x$. On the other hand, if $beta^k(x)neq x$, then it must be the case that $beta(x)neq x$ (since $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.
answered yesterday
Mike EarnestMike Earnest
24.3k22151
$endgroup$
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
add a comment |
$begingroup$
The statement to be proved is untrue under your definition of disjoint:
Your Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)=x$ implies $beta(x)neq x$ and vice versa.
To see this, consider $alpha=(1;2)(3)(4)$ and $beta=(1)(2)(3;4)$, $k=2$.
You get a correct result is you instead use
Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)neq x$ implies $beta(x)= x$ and vice versa.
To see this, suppose $alpha(x)neq x$. Then disjointness implies $beta(x)=x$, so $beta^k(x)=beta^{k-1}(x)=dots=beta(x)=x$. On the other hand, if $beta^k(x)neq x$, then it must be the case that $beta(x)neq x$ (since $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.
answered yesterday
Mike EarnestMike Earnest
24.3k22151
$endgroup$
The statement to be proved is untrue under your definition of disjoint:
Your Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)=x$ implies $beta(x)neq x$ and vice versa.
To see this, consider $alpha=(1;2)(3)(4)$ and $beta=(1)(2)(3;4)$, $k=2$.
You get a correct result is you instead use
Defintion (Disjoint): $alpha$ and $beta$ are disjoint if $alpha(x)neq x$ implies $beta(x)= x$ and vice versa.
To see this, suppose $alpha(x)neq x$. Then disjointness implies $beta(x)=x$, so $beta^k(x)=beta^{k-1}(x)=dots=beta(x)=x$. On the other hand, if $beta^k(x)neq x$, then it must be the case that $beta(x)neq x$ (since $beta(x)=ximplies beta^k(x)=x$ as shown before), so disjointness implies $alpha(x)=x$. This proves $alpha$ and $beta^k$ are disjoint.
answered yesterday
Mike EarnestMike Earnest
24.3k22151
answered yesterday
Mike EarnestMike Earnest
24.3k22151
answered yesterday
Mike EarnestMike Earnest
24.3k22151
answered yesterday
Mike EarnestMike Earnest
24.3k22151
24.3k22151
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
add a comment |
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
$begingroup$
Downvoter, care to explain?
$endgroup$
– Mike Earnest
yesterday
add a comment |
$begingroup$
The set of elements that is moved by $beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $alpha$ (let's call it $A$) so if you apply $beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.
answered yesterday
gammagamma
3,39221342
$endgroup$
add a comment |
$begingroup$
The set of elements that is moved by $beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $alpha$ (let's call it $A$) so if you apply $beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.
answered yesterday
gammagamma
3,39221342
$endgroup$
add a comment |
$begingroup$
The set of elements that is moved by $beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $alpha$ (let's call it $A$) so if you apply $beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.
answered yesterday
gammagamma
3,39221342
$endgroup$
The set of elements that is moved by $beta$ (let's call it $B$) is disjoint with the set of elements that is moved by $alpha$ (let's call it $A$) so if you apply $beta$ total of $k$ times, you still only move things in $B$, that is disjoint from $A$.
answered yesterday
gammagamma
3,39221342
answered yesterday
gammagamma
3,39221342
answered yesterday
gammagamma
3,39221342
answered yesterday
gammagamma
3,39221342
3,39221342
add a comment |
add a comment |
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