Mixing
Finding the center and normal subgroup of $D_{10}$
$begingroup$
Consider $D_{10}$, the dihedral group of order $10$. This is the group of symmetries of the regular pentagon. Label the vertices of the pentagon clockwise as $1,2,3,4,5$. Let $sigma = (12345)$ and $tau = (13)(45)$.
How do I find the center and normal subgroup of $D_{10}$?
I have subgroups of $D_{10}$, but I am not sure how that will help me to find the normal subgroup of $D_{10}$.
Please help me.
abstract-algebra group-theory normal-subgroups
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
$endgroup$
add a comment |
$begingroup$
Consider $D_{10}$, the dihedral group of order $10$. This is the group of symmetries of the regular pentagon. Label the vertices of the pentagon clockwise as $1,2,3,4,5$. Let $sigma = (12345)$ and $tau = (13)(45)$.
How do I find the center and normal subgroup of $D_{10}$?
I have subgroups of $D_{10}$, but I am not sure how that will help me to find the normal subgroup of $D_{10}$.
Please help me.
abstract-algebra group-theory normal-subgroups
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
$endgroup$
$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
1
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45
add a comment |
$begingroup$
Consider $D_{10}$, the dihedral group of order $10$. This is the group of symmetries of the regular pentagon. Label the vertices of the pentagon clockwise as $1,2,3,4,5$. Let $sigma = (12345)$ and $tau = (13)(45)$.
How do I find the center and normal subgroup of $D_{10}$?
I have subgroups of $D_{10}$, but I am not sure how that will help me to find the normal subgroup of $D_{10}$.
Please help me.
abstract-algebra group-theory normal-subgroups
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
$endgroup$
Consider $D_{10}$, the dihedral group of order $10$. This is the group of symmetries of the regular pentagon. Label the vertices of the pentagon clockwise as $1,2,3,4,5$. Let $sigma = (12345)$ and $tau = (13)(45)$.
How do I find the center and normal subgroup of $D_{10}$?
I have subgroups of $D_{10}$, but I am not sure how that will help me to find the normal subgroup of $D_{10}$.
Please help me.
abstract-algebra group-theory normal-subgroups
abstract-algebra group-theory normal-subgroups
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
edited Nov 27 '15 at 0:21
Silvia Ghinassi
3,89671641
3,89671641
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
asked Nov 26 '15 at 23:35
wannnhikmahwannnhikmah
61
61
$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
1
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45
add a comment |
$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
1
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45
$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
1
1
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.
$D_{10}$ has the presentation
$$D_{10} = left<r,f : r^5 = f^2 = e, rf = fr^{-1}right>.$$
Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 leq a leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just ${e}$.
Now for normal subgroups. If you include any element $r^a$ for $1 leq a leq 4$, you get all of $left<rright>$. So then if you also add any of the elements $r^af$, where $0 leq a leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 leq a,b leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $left<rright>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are ${e}, D_{10}, left<rright>$, and $left<r^afright>$ for each $0 leq a leq 4$. Only the first three are normal ($left<rright>$ is normal because it has index $2$). You can easily show $left<r^afright>$ is not normal by multiplying on the left and right by $f$ if $a neq 0$ and by $r$ if $a = 0$.
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
add a comment |
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$begingroup$
One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.
$D_{10}$ has the presentation
$$D_{10} = left<r,f : r^5 = f^2 = e, rf = fr^{-1}right>.$$
Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 leq a leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just ${e}$.
Now for normal subgroups. If you include any element $r^a$ for $1 leq a leq 4$, you get all of $left<rright>$. So then if you also add any of the elements $r^af$, where $0 leq a leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 leq a,b leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $left<rright>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are ${e}, D_{10}, left<rright>$, and $left<r^afright>$ for each $0 leq a leq 4$. Only the first three are normal ($left<rright>$ is normal because it has index $2$). You can easily show $left<r^afright>$ is not normal by multiplying on the left and right by $f$ if $a neq 0$ and by $r$ if $a = 0$.
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
add a comment |
$begingroup$
One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.
$D_{10}$ has the presentation
$$D_{10} = left<r,f : r^5 = f^2 = e, rf = fr^{-1}right>.$$
Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 leq a leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just ${e}$.
Now for normal subgroups. If you include any element $r^a$ for $1 leq a leq 4$, you get all of $left<rright>$. So then if you also add any of the elements $r^af$, where $0 leq a leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 leq a,b leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $left<rright>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are ${e}, D_{10}, left<rright>$, and $left<r^afright>$ for each $0 leq a leq 4$. Only the first three are normal ($left<rright>$ is normal because it has index $2$). You can easily show $left<r^afright>$ is not normal by multiplying on the left and right by $f$ if $a neq 0$ and by $r$ if $a = 0$.
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
add a comment |
$begingroup$
One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.
$D_{10}$ has the presentation
$$D_{10} = left<r,f : r^5 = f^2 = e, rf = fr^{-1}right>.$$
Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 leq a leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just ${e}$.
Now for normal subgroups. If you include any element $r^a$ for $1 leq a leq 4$, you get all of $left<rright>$. So then if you also add any of the elements $r^af$, where $0 leq a leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 leq a,b leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $left<rright>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are ${e}, D_{10}, left<rright>$, and $left<r^afright>$ for each $0 leq a leq 4$. Only the first three are normal ($left<rright>$ is normal because it has index $2$). You can easily show $left<r^afright>$ is not normal by multiplying on the left and right by $f$ if $a neq 0$ and by $r$ if $a = 0$.
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
$endgroup$
One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.
$D_{10}$ has the presentation
$$D_{10} = left<r,f : r^5 = f^2 = e, rf = fr^{-1}right>.$$
Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 leq a leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just ${e}$.
Now for normal subgroups. If you include any element $r^a$ for $1 leq a leq 4$, you get all of $left<rright>$. So then if you also add any of the elements $r^af$, where $0 leq a leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 leq a,b leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $left<rright>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are ${e}, D_{10}, left<rright>$, and $left<r^afright>$ for each $0 leq a leq 4$. Only the first three are normal ($left<rright>$ is normal because it has index $2$). You can easily show $left<r^afright>$ is not normal by multiplying on the left and right by $f$ if $a neq 0$ and by $r$ if $a = 0$.
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
answered Nov 27 '15 at 1:07
Ethan AlwaiseEthan Alwaise
6,193617
6,193617
add a comment |
add a comment |
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$begingroup$
An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$.
$endgroup$
– Franciele Daltoé
Nov 27 '15 at 0:44
1
$begingroup$
It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal.
$endgroup$
– Ethan Alwaise
Nov 27 '15 at 0:45