$S_3oplus S_3$ has an element of order $4,6,9$ or $18$.
$S_3oplus S_3$ has an element of order $4,6,9$ or $18$. $oplus$ is the direct sum.
Its order is $36$ so by sylow theorem it has subgroups of order $4$ and $9$. Also order of a permutation is the lcm of the order of disjoint cycles. But when I consider elements of $S_3oplus S_3$ the cycles might not be disjoint eg. $(12)(123)$. It says the answer is 6 but I'm not convinced because of this reason.
group-theory permutations least-common-multiple
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
add a comment |
$S_3oplus S_3$ has an element of order $4,6,9$ or $18$. $oplus$ is the direct sum.
Its order is $36$ so by sylow theorem it has subgroups of order $4$ and $9$. Also order of a permutation is the lcm of the order of disjoint cycles. But when I consider elements of $S_3oplus S_3$ the cycles might not be disjoint eg. $(12)(123)$. It says the answer is 6 but I'm not convinced because of this reason.
group-theory permutations least-common-multiple
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
1
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14
add a comment |
$S_3oplus S_3$ has an element of order $4,6,9$ or $18$. $oplus$ is the direct sum.
Its order is $36$ so by sylow theorem it has subgroups of order $4$ and $9$. Also order of a permutation is the lcm of the order of disjoint cycles. But when I consider elements of $S_3oplus S_3$ the cycles might not be disjoint eg. $(12)(123)$. It says the answer is 6 but I'm not convinced because of this reason.
group-theory permutations least-common-multiple
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
$S_3oplus S_3$ has an element of order $4,6,9$ or $18$. $oplus$ is the direct sum.
Its order is $36$ so by sylow theorem it has subgroups of order $4$ and $9$. Also order of a permutation is the lcm of the order of disjoint cycles. But when I consider elements of $S_3oplus S_3$ the cycles might not be disjoint eg. $(12)(123)$. It says the answer is 6 but I'm not convinced because of this reason.
group-theory permutations least-common-multiple
group-theory permutations least-common-multiple
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
edited Jan 4 at 10:48
Rhaldryn
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
asked Jan 4 at 10:37
RhaldrynRhaldryn
309414
309414
I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
1
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14
add a comment |
I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
1
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14
I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
1
1
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14
add a comment |
1 Answer
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That $oplus$ symbol? It's the direct sum; anything from one piece commutes with anything from the other. We can represent the elements as ordered pairs $(a,b)$, with group operation $(a,b)cdot (c,d) = (acdot c, bcdot d)$. It doesn't matter if the cycles aren't disjoint between the two objects - we're not going to multiply across that line anyway.
answered Jan 4 at 10:45
jmerryjmerry
2,511312
add a comment |
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1 Answer
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That $oplus$ symbol? It's the direct sum; anything from one piece commutes with anything from the other. We can represent the elements as ordered pairs $(a,b)$, with group operation $(a,b)cdot (c,d) = (acdot c, bcdot d)$. It doesn't matter if the cycles aren't disjoint between the two objects - we're not going to multiply across that line anyway.
answered Jan 4 at 10:45
jmerryjmerry
2,511312
add a comment |
That $oplus$ symbol? It's the direct sum; anything from one piece commutes with anything from the other. We can represent the elements as ordered pairs $(a,b)$, with group operation $(a,b)cdot (c,d) = (acdot c, bcdot d)$. It doesn't matter if the cycles aren't disjoint between the two objects - we're not going to multiply across that line anyway.
answered Jan 4 at 10:45
jmerryjmerry
2,511312
add a comment |
That $oplus$ symbol? It's the direct sum; anything from one piece commutes with anything from the other. We can represent the elements as ordered pairs $(a,b)$, with group operation $(a,b)cdot (c,d) = (acdot c, bcdot d)$. It doesn't matter if the cycles aren't disjoint between the two objects - we're not going to multiply across that line anyway.
answered Jan 4 at 10:45
jmerryjmerry
2,511312
That $oplus$ symbol? It's the direct sum; anything from one piece commutes with anything from the other. We can represent the elements as ordered pairs $(a,b)$, with group operation $(a,b)cdot (c,d) = (acdot c, bcdot d)$. It doesn't matter if the cycles aren't disjoint between the two objects - we're not going to multiply across that line anyway.
answered Jan 4 at 10:45
jmerryjmerry
2,511312
answered Jan 4 at 10:45
jmerryjmerry
2,511312
answered Jan 4 at 10:45
jmerryjmerry
2,511312
answered Jan 4 at 10:45
jmerryjmerry
2,511312
2,511312
add a comment |
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I guess $oplus$ is supposed to be the direct product?
– 0x539
Jan 4 at 10:44
@0x539 Or possibly direct sum. Not that it makes much difference in this case.
– Arthur
Jan 4 at 10:46
1
The first and second $S_3$ have nothing to do with each other. To prevent confusion, it could probably help to see $S_3 oplus S_3$ as a subgroup of $S_6$, where the first $S_3$ is $S({1,2,3})$ and the second is $S({4,5,6})$.
– TastyRomeo
Jan 4 at 10:59
@TastyRomeo Yeah that was the source of the confusion, ty.
– Rhaldryn
Jan 4 at 15:14