Is $S_10$ generated by the group of $6$ cycles?

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Is $S_10$ generated by the subgroup of $6$ cycles?




I'll denote that subgroup by $H$.



We know that by conjugation permutations remains with the same cycle structure, and therefore $H$ is closed to conjugation. It is easy to see from here that $H$ is normal subgroup.



The only normal subgroup of $S_10$ is $A_10$, therefore $H$ must be either $S_10$ or $A_10$.



How can i determine which one from here?



Are the specific numbers $10,6$ even relevant?






group-theory permutations normal-subgroups







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  • 2




    Hint: what is the sign of a $6$-cycle?
    – Wojowu
    Sep 4 at 10:38






  • 5




    The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
    – Derek Holt
    Sep 4 at 10:39














up vote
3
down vote

favorite













Is $S_10$ generated by the subgroup of $6$ cycles?




I'll denote that subgroup by $H$.



We know that by conjugation permutations remains with the same cycle structure, and therefore $H$ is closed to conjugation. It is easy to see from here that $H$ is normal subgroup.



The only normal subgroup of $S_10$ is $A_10$, therefore $H$ must be either $S_10$ or $A_10$.



How can i determine which one from here?



Are the specific numbers $10,6$ even relevant?






group-theory permutations normal-subgroups







share|cite|improve this question

















  • 2




    Hint: what is the sign of a $6$-cycle?
    – Wojowu
    Sep 4 at 10:38






  • 5




    The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
    – Derek Holt
    Sep 4 at 10:39












up vote
3
down vote

favorite









up vote
3
down vote

favorite












Is $S_10$ generated by the subgroup of $6$ cycles?




I'll denote that subgroup by $H$.



We know that by conjugation permutations remains with the same cycle structure, and therefore $H$ is closed to conjugation. It is easy to see from here that $H$ is normal subgroup.



The only normal subgroup of $S_10$ is $A_10$, therefore $H$ must be either $S_10$ or $A_10$.



How can i determine which one from here?



Are the specific numbers $10,6$ even relevant?






group-theory permutations normal-subgroups







share|cite|improve this question














Is $S_10$ generated by the subgroup of $6$ cycles?




I'll denote that subgroup by $H$.



We know that by conjugation permutations remains with the same cycle structure, and therefore $H$ is closed to conjugation. It is easy to see from here that $H$ is normal subgroup.



The only normal subgroup of $S_10$ is $A_10$, therefore $H$ must be either $S_10$ or $A_10$.



How can i determine which one from here?



Are the specific numbers $10,6$ even relevant?






group-theory permutations normal-subgroups




group-theory permutations normal-subgroups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 4 at 10:34









ChikChak

712217




712217







  • 2




    Hint: what is the sign of a $6$-cycle?
    – Wojowu
    Sep 4 at 10:38






  • 5




    The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
    – Derek Holt
    Sep 4 at 10:39












  • 2




    Hint: what is the sign of a $6$-cycle?
    – Wojowu
    Sep 4 at 10:38






  • 5




    The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
    – Derek Holt
    Sep 4 at 10:39







2




2




Hint: what is the sign of a $6$-cycle?
– Wojowu
Sep 4 at 10:38




Hint: what is the sign of a $6$-cycle?
– Wojowu
Sep 4 at 10:38




5




5




The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
– Derek Holt
Sep 4 at 10:39




The set of $6$-cycles do not form a subgroup. You mean is it generated by the subset of all $6$-cycles.
– Derek Holt
Sep 4 at 10:39










1 Answer
1






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oldest

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up vote
4
down vote



accepted










Each $6$-cycle has sign equal to $-1$. Therefore, $Hnotsubset A_10$.






share|cite|improve this answer




















  • Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
    – ChikChak
    Sep 4 at 10:49






  • 1




    @ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
    – José Carlos Santos
    Sep 4 at 10:51










  • Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
    – ChikChak
    Sep 4 at 10:55






  • 1




    @ChikChak No. For instance, $ein H$ and it is an even permutation.
    – José Carlos Santos
    Sep 4 at 10:57










  • So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
    – ChikChak
    Sep 4 at 11:03










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Each $6$-cycle has sign equal to $-1$. Therefore, $Hnotsubset A_10$.






share|cite|improve this answer




















  • Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
    – ChikChak
    Sep 4 at 10:49






  • 1




    @ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
    – José Carlos Santos
    Sep 4 at 10:51










  • Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
    – ChikChak
    Sep 4 at 10:55






  • 1




    @ChikChak No. For instance, $ein H$ and it is an even permutation.
    – José Carlos Santos
    Sep 4 at 10:57










  • So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
    – ChikChak
    Sep 4 at 11:03














up vote
4
down vote



accepted










Each $6$-cycle has sign equal to $-1$. Therefore, $Hnotsubset A_10$.






share|cite|improve this answer




















  • Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
    – ChikChak
    Sep 4 at 10:49






  • 1




    @ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
    – José Carlos Santos
    Sep 4 at 10:51










  • Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
    – ChikChak
    Sep 4 at 10:55






  • 1




    @ChikChak No. For instance, $ein H$ and it is an even permutation.
    – José Carlos Santos
    Sep 4 at 10:57










  • So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
    – ChikChak
    Sep 4 at 11:03












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Each $6$-cycle has sign equal to $-1$. Therefore, $Hnotsubset A_10$.






share|cite|improve this answer












Each $6$-cycle has sign equal to $-1$. Therefore, $Hnotsubset A_10$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 4 at 10:38









José Carlos Santos

122k16101186




122k16101186











  • Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
    – ChikChak
    Sep 4 at 10:49






  • 1




    @ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
    – José Carlos Santos
    Sep 4 at 10:51










  • Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
    – ChikChak
    Sep 4 at 10:55






  • 1




    @ChikChak No. For instance, $ein H$ and it is an even permutation.
    – José Carlos Santos
    Sep 4 at 10:57










  • So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
    – ChikChak
    Sep 4 at 11:03
















  • Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
    – ChikChak
    Sep 4 at 10:49






  • 1




    @ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
    – José Carlos Santos
    Sep 4 at 10:51










  • Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
    – ChikChak
    Sep 4 at 10:55






  • 1




    @ChikChak No. For instance, $ein H$ and it is an even permutation.
    – José Carlos Santos
    Sep 4 at 10:57










  • So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
    – ChikChak
    Sep 4 at 11:03















Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
– ChikChak
Sep 4 at 10:49




Wow, so it's much simpler than I thought, because it just doesn't contain any even permutation, it obviously can't be whole $S_10$.... Correct?
– ChikChak
Sep 4 at 10:49




1




1




@ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
– José Carlos Santos
Sep 4 at 10:51




@ChikChak Yes, that's correct, if the “it” means the set of all $6$-cycles.
– José Carlos Santos
Sep 4 at 10:51












Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
– ChikChak
Sep 4 at 10:55




Carlos Santos I mean the group generated by the set of all the $6$-cycles. Is it correct?
– ChikChak
Sep 4 at 10:55




1




1




@ChikChak No. For instance, $ein H$ and it is an even permutation.
– José Carlos Santos
Sep 4 at 10:57




@ChikChak No. For instance, $ein H$ and it is an even permutation.
– José Carlos Santos
Sep 4 at 10:57












So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
– ChikChak
Sep 4 at 11:03




So $H$ must be whole $S_10$, right? As it is normal, and isn't $A_10$ because it contains also odd permutations.
– ChikChak
Sep 4 at 11:03

















 

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