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The normalizer subgroup of $(),(12)(34)$
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Let $H$ be the subgroup $(),(12)(34)$ of $S_4$. Find the normalizer $N(H)$ and the quotient $N(H)$.
What I have so far: in this case, $N(H)=Z((12)(34))$. The order of the conjugacy class of $(12)(34)$ is $3$, so the order of its centralizer (which is $N(H)$) is $8$. The quotient has order $4$ so it's either $C_2times C_2$ or $C_4$.
To find what elements lie in $N(H)$, do I need to check whether each of the $24$ elements commute with $(12)(34)$ (until I find $8$ elements), or is there an easy way to see what elements lie in $N(H)$? How to understand whether the quotient is $C_4$ or $C_2times C_2$?
abstract-algebra group-theory permutations
asked Aug 5 at 23:46
user437309
556212
add a comment |Â
up vote
3
down vote
favorite
Let $H$ be the subgroup $(),(12)(34)$ of $S_4$. Find the normalizer $N(H)$ and the quotient $N(H)$.
What I have so far: in this case, $N(H)=Z((12)(34))$. The order of the conjugacy class of $(12)(34)$ is $3$, so the order of its centralizer (which is $N(H)$) is $8$. The quotient has order $4$ so it's either $C_2times C_2$ or $C_4$.
To find what elements lie in $N(H)$, do I need to check whether each of the $24$ elements commute with $(12)(34)$ (until I find $8$ elements), or is there an easy way to see what elements lie in $N(H)$? How to understand whether the quotient is $C_4$ or $C_2times C_2$?
abstract-algebra group-theory permutations
asked Aug 5 at 23:46
user437309
556212
I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $H$ be the subgroup $(),(12)(34)$ of $S_4$. Find the normalizer $N(H)$ and the quotient $N(H)$.
What I have so far: in this case, $N(H)=Z((12)(34))$. The order of the conjugacy class of $(12)(34)$ is $3$, so the order of its centralizer (which is $N(H)$) is $8$. The quotient has order $4$ so it's either $C_2times C_2$ or $C_4$.
To find what elements lie in $N(H)$, do I need to check whether each of the $24$ elements commute with $(12)(34)$ (until I find $8$ elements), or is there an easy way to see what elements lie in $N(H)$? How to understand whether the quotient is $C_4$ or $C_2times C_2$?
abstract-algebra group-theory permutations
asked Aug 5 at 23:46
user437309
556212
Let $H$ be the subgroup $(),(12)(34)$ of $S_4$. Find the normalizer $N(H)$ and the quotient $N(H)$.
What I have so far: in this case, $N(H)=Z((12)(34))$. The order of the conjugacy class of $(12)(34)$ is $3$, so the order of its centralizer (which is $N(H)$) is $8$. The quotient has order $4$ so it's either $C_2times C_2$ or $C_4$.
To find what elements lie in $N(H)$, do I need to check whether each of the $24$ elements commute with $(12)(34)$ (until I find $8$ elements), or is there an easy way to see what elements lie in $N(H)$? How to understand whether the quotient is $C_4$ or $C_2times C_2$?
abstract-algebra group-theory permutations
asked Aug 5 at 23:46
user437309
556212
asked Aug 5 at 23:46
user437309
556212
asked Aug 5 at 23:46
user437309
556212
asked Aug 5 at 23:46
user437309
556212
556212
I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53
add a comment |Â
I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53
I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^-1 = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H cong C_2 times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$
answered Aug 6 at 1:17
Stefan4024
28.5k53174
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^-1 = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H cong C_2 times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$
answered Aug 6 at 1:17
Stefan4024
28.5k53174
add a comment |Â
up vote
2
down vote
accepted
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^-1 = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H cong C_2 times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$
answered Aug 6 at 1:17
Stefan4024
28.5k53174
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^-1 = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H cong C_2 times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$
answered Aug 6 at 1:17
Stefan4024
28.5k53174
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^-1 = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H cong C_2 times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$
answered Aug 6 at 1:17
Stefan4024
28.5k53174
answered Aug 6 at 1:17
Stefan4024
28.5k53174
answered Aug 6 at 1:17
Stefan4024
28.5k53174
answered Aug 6 at 1:17
Stefan4024
28.5k53174
28.5k53174
add a comment |Â
add a comment |Â
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I presume that $()$ is supposed to be the identity, but I've never seen that notation for it.
– Cameron Buie
Aug 5 at 23:49
Yes, it's supposed to be the identity.
– user437309
Aug 6 at 1:06
@CameronBuie This notation for the identity is used e.g. here groupprops.subwiki.org/wiki/… (I believe that's where I learned it from).
– user437309
Aug 6 at 15:53