How many cyclic subgroups are in the Dihedral group?
$begingroup$
let $D_8$ be the group of symmetries of a square. Not counting the trivial subgroup, how many distinct cyclic subgroup s does $D_8$ contain?
My result:
A group $D_8$, contains an element $a$ of order 4.
$<{a}> =(a^4:4in Z)$
$(1.a,a^2,a^3,x,b,c,d)$ the reflections and symmetries in $D_8$.
$(1)(a)=a
\
(a)(a)=a^2
\
(a)(a^2)=a^3
\
(a)(a^3)=a^4=1$
$(1,a,a^2,a^3)$
Did I answer this correctly? If yes can someone clarify my writing or if it's wrong please guide me ...
abstract-algebra group-theory cyclic-groups
edited Oct 8 '13 at 15:11
user1729
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
$endgroup$
add a comment |
$begingroup$
let $D_8$ be the group of symmetries of a square. Not counting the trivial subgroup, how many distinct cyclic subgroup s does $D_8$ contain?
My result:
A group $D_8$, contains an element $a$ of order 4.
$<{a}> =(a^4:4in Z)$
$(1.a,a^2,a^3,x,b,c,d)$ the reflections and symmetries in $D_8$.
$(1)(a)=a
\
(a)(a)=a^2
\
(a)(a^2)=a^3
\
(a)(a^3)=a^4=1$
$(1,a,a^2,a^3)$
Did I answer this correctly? If yes can someone clarify my writing or if it's wrong please guide me ...
abstract-algebra group-theory cyclic-groups
edited Oct 8 '13 at 15:11
user1729
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
$endgroup$
add a comment |
$begingroup$
let $D_8$ be the group of symmetries of a square. Not counting the trivial subgroup, how many distinct cyclic subgroup s does $D_8$ contain?
My result:
A group $D_8$, contains an element $a$ of order 4.
$<{a}> =(a^4:4in Z)$
$(1.a,a^2,a^3,x,b,c,d)$ the reflections and symmetries in $D_8$.
$(1)(a)=a
\
(a)(a)=a^2
\
(a)(a^2)=a^3
\
(a)(a^3)=a^4=1$
$(1,a,a^2,a^3)$
Did I answer this correctly? If yes can someone clarify my writing or if it's wrong please guide me ...
abstract-algebra group-theory cyclic-groups
edited Oct 8 '13 at 15:11
user1729
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
$endgroup$
let $D_8$ be the group of symmetries of a square. Not counting the trivial subgroup, how many distinct cyclic subgroup s does $D_8$ contain?
My result:
A group $D_8$, contains an element $a$ of order 4.
$<{a}> =(a^4:4in Z)$
$(1.a,a^2,a^3,x,b,c,d)$ the reflections and symmetries in $D_8$.
$(1)(a)=a
\
(a)(a)=a^2
\
(a)(a^2)=a^3
\
(a)(a^3)=a^4=1$
$(1,a,a^2,a^3)$
Did I answer this correctly? If yes can someone clarify my writing or if it's wrong please guide me ...
abstract-algebra group-theory cyclic-groups
abstract-algebra group-theory cyclic-groups
edited Oct 8 '13 at 15:11
user1729
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
edited Oct 8 '13 at 15:11
user1729
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
edited Oct 8 '13 at 15:11
user1729
17k64085
edited Oct 8 '13 at 15:11
user1729
17k64085
edited Oct 8 '13 at 15:11
user1729
17k64085
17k64085
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
asked Sep 22 '13 at 22:53
miguel barnesmiguel barnes
13917
13917
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$${1, a^2}$$
Four additional cyclic subgroups of order two are as follows:
$${1, x}, {1, b}, {1, c}, {1, d}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: ${1}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
answered Sep 22 '13 at 23:01
amWhyamWhy
1
$endgroup$
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
add a comment |
$begingroup$
Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup.
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer.
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
$endgroup$
add a comment |
$begingroup$
Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
f:=FreeGroup("a","b");;
a:=f.1;; b:=f.2;;
s:=f/[a^2,b^4,(a*b)^2];
e:=Subgroups(s);;
for i in [1..8] do Print(e[i]," ",IsCyclic(e[i]),"n"); od;
Group( <identity ...> ) true
Group( [ b^2 ] ) true
Group( [ a ] ) true
Group( [ b^-1*a*b ] ) true
Group( [ a*b ] ) true
Group( [ b*a ] ) true
Group( [ a, b^-2 ] ) false
Group( [ b ] ) true
answered Oct 8 '13 at 15:04
mrsmrs
1
$endgroup$
$begingroup$
Length ofeis 10, not 8 - any considerations against using more genericfor i in [1..Length(e)]syntax?
$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
add a comment |
$begingroup$
No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$) is
$tau{(n)} +$ $n$. Where $tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $tau(n) + sigma(n)$ where $sigma(n)$ is sum of all positive divisor.
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$${1, a^2}$$
Four additional cyclic subgroups of order two are as follows:
$${1, x}, {1, b}, {1, c}, {1, d}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: ${1}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
answered Sep 22 '13 at 23:01
amWhyamWhy
1
$endgroup$
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
add a comment |
$begingroup$
You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$${1, a^2}$$
Four additional cyclic subgroups of order two are as follows:
$${1, x}, {1, b}, {1, c}, {1, d}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: ${1}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
answered Sep 22 '13 at 23:01
amWhyamWhy
1
$endgroup$
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
add a comment |
$begingroup$
You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$${1, a^2}$$
Four additional cyclic subgroups of order two are as follows:
$${1, x}, {1, b}, {1, c}, {1, d}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: ${1}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
answered Sep 22 '13 at 23:01
amWhyamWhy
1
$endgroup$
You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$${1, a^2}$$
Four additional cyclic subgroups of order two are as follows:
$${1, x}, {1, b}, {1, c}, {1, d}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: ${1}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
answered Sep 22 '13 at 23:01
amWhyamWhy
1
edited Sep 23 '13 at 1:59
answered Sep 22 '13 at 23:01
amWhyamWhy
1
answered Sep 22 '13 at 23:01
amWhyamWhy
1
answered Sep 22 '13 at 23:01
amWhyamWhy
1
1
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
add a comment |
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
$begingroup$
How did you obtain those cyclic subgroups? Isn't $(1,x)$ a cyclic subgroup as well?
$endgroup$
– miguel barnes
Sep 23 '13 at 1:56
1
1
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
$begingroup$
Yes, ${1, x}$ is another cyclic subgroup of order 2. I missed seeing x in your set of elements.
$endgroup$
– amWhy
Sep 23 '13 at 1:58
add a comment |
$begingroup$
Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup.
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer.
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
$endgroup$
add a comment |
$begingroup$
Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup.
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer.
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
$endgroup$
add a comment |
$begingroup$
Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup.
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer.
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
$endgroup$
Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup.
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer.
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
answered Sep 22 '13 at 23:02
Ittay WeissIttay Weiss
63.8k6101184
63.8k6101184
add a comment |
add a comment |
$begingroup$
Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
f:=FreeGroup("a","b");;
a:=f.1;; b:=f.2;;
s:=f/[a^2,b^4,(a*b)^2];
e:=Subgroups(s);;
for i in [1..8] do Print(e[i]," ",IsCyclic(e[i]),"n"); od;
Group( <identity ...> ) true
Group( [ b^2 ] ) true
Group( [ a ] ) true
Group( [ b^-1*a*b ] ) true
Group( [ a*b ] ) true
Group( [ b*a ] ) true
Group( [ a, b^-2 ] ) false
Group( [ b ] ) true
answered Oct 8 '13 at 15:04
mrsmrs
1
$endgroup$
$begingroup$
Length ofeis 10, not 8 - any considerations against using more genericfor i in [1..Length(e)]syntax?
$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
add a comment |
$begingroup$
Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
f:=FreeGroup("a","b");;
a:=f.1;; b:=f.2;;
s:=f/[a^2,b^4,(a*b)^2];
e:=Subgroups(s);;
for i in [1..8] do Print(e[i]," ",IsCyclic(e[i]),"n"); od;
Group( <identity ...> ) true
Group( [ b^2 ] ) true
Group( [ a ] ) true
Group( [ b^-1*a*b ] ) true
Group( [ a*b ] ) true
Group( [ b*a ] ) true
Group( [ a, b^-2 ] ) false
Group( [ b ] ) true
answered Oct 8 '13 at 15:04
mrsmrs
1
$endgroup$
$begingroup$
Length ofeis 10, not 8 - any considerations against using more genericfor i in [1..Length(e)]syntax?
$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
add a comment |
$begingroup$
Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
f:=FreeGroup("a","b");;
a:=f.1;; b:=f.2;;
s:=f/[a^2,b^4,(a*b)^2];
e:=Subgroups(s);;
for i in [1..8] do Print(e[i]," ",IsCyclic(e[i]),"n"); od;
Group( <identity ...> ) true
Group( [ b^2 ] ) true
Group( [ a ] ) true
Group( [ b^-1*a*b ] ) true
Group( [ a*b ] ) true
Group( [ b*a ] ) true
Group( [ a, b^-2 ] ) false
Group( [ b ] ) true
answered Oct 8 '13 at 15:04
mrsmrs
1
$endgroup$
Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
f:=FreeGroup("a","b");;
a:=f.1;; b:=f.2;;
s:=f/[a^2,b^4,(a*b)^2];
e:=Subgroups(s);;
for i in [1..8] do Print(e[i]," ",IsCyclic(e[i]),"n"); od;
Group( <identity ...> ) true
Group( [ b^2 ] ) true
Group( [ a ] ) true
Group( [ b^-1*a*b ] ) true
Group( [ a*b ] ) true
Group( [ b*a ] ) true
Group( [ a, b^-2 ] ) false
Group( [ b ] ) true
answered Oct 8 '13 at 15:04
mrsmrs
1
edited Oct 8 '13 at 15:10
answered Oct 8 '13 at 15:04
mrsmrs
1
answered Oct 8 '13 at 15:04
mrsmrs
1
answered Oct 8 '13 at 15:04
mrsmrs
1
1
$begingroup$
Length ofeis 10, not 8 - any considerations against using more genericfor i in [1..Length(e)]syntax?
$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
add a comment |
$begingroup$
Length ofeis 10, not 8 - any considerations against using more genericfor i in [1..Length(e)]syntax?
$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
$begingroup$
Length of
e is 10, not 8 - any considerations against using more generic for i in [1..Length(e)] syntax?$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
$begingroup$
Length of
e is 10, not 8 - any considerations against using more generic for i in [1..Length(e)] syntax?$endgroup$
– Alexander Konovalov
Oct 8 '13 at 23:24
add a comment |
$begingroup$
No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$) is
$tau{(n)} +$ $n$. Where $tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $tau(n) + sigma(n)$ where $sigma(n)$ is sum of all positive divisor.
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
$endgroup$
add a comment |
$begingroup$
No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$) is
$tau{(n)} +$ $n$. Where $tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $tau(n) + sigma(n)$ where $sigma(n)$ is sum of all positive divisor.
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
$endgroup$
add a comment |
$begingroup$
No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$) is
$tau{(n)} +$ $n$. Where $tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $tau(n) + sigma(n)$ where $sigma(n)$ is sum of all positive divisor.
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
$endgroup$
No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$) is
$tau{(n)} +$ $n$. Where $tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $tau(n) + sigma(n)$ where $sigma(n)$ is sum of all positive divisor.
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
edited Dec 11 '18 at 9:39
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
answered Dec 11 '18 at 9:34
August ariesAugust aries
13
13
add a comment |
add a comment |
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