Determine number of conjugacy class in $D_8$
I’m using the formula that the number of conjugacy class is given to be $frac{1}{|G|}sum|C_{G}(g)|$, where $C_{G}(g)={h in G ; gh=hg}$, which is a special result by Burnside’s theorem.
I found that the number of conjugacy class in $D_8$ is 5, so to double check, I listed down all $C_{G}(g)={h in G ; gh=hg}$.
Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.
Well, $|C_G(1)|=|D_8|=16$,
$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7}|=8$,
$|C_G(r^4)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m}|=9$ since $r^4m=mr^4$
Similarly, we can count for the reflections;
$|C_G(m)|=|C_G(r^4m)|=|{1,m}|=2$, while the rest of the elements with any reflections only has one element, i.e $|C_G(r^nm)|=|{1}|=1,n neq 0,4$
So the problem comes that my summation is 83, while my $|G|=16$. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...
abstract-algebra group-theory finite-groups group-actions dihedral-groups
edited Nov 27 '18 at 4:17
Shubham
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
add a comment |
I’m using the formula that the number of conjugacy class is given to be $frac{1}{|G|}sum|C_{G}(g)|$, where $C_{G}(g)={h in G ; gh=hg}$, which is a special result by Burnside’s theorem.
I found that the number of conjugacy class in $D_8$ is 5, so to double check, I listed down all $C_{G}(g)={h in G ; gh=hg}$.
Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.
Well, $|C_G(1)|=|D_8|=16$,
$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7}|=8$,
$|C_G(r^4)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m}|=9$ since $r^4m=mr^4$
Similarly, we can count for the reflections;
$|C_G(m)|=|C_G(r^4m)|=|{1,m}|=2$, while the rest of the elements with any reflections only has one element, i.e $|C_G(r^nm)|=|{1}|=1,n neq 0,4$
So the problem comes that my summation is 83, while my $|G|=16$. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...
abstract-algebra group-theory finite-groups group-actions dihedral-groups
edited Nov 27 '18 at 4:17
Shubham
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10
add a comment |
I’m using the formula that the number of conjugacy class is given to be $frac{1}{|G|}sum|C_{G}(g)|$, where $C_{G}(g)={h in G ; gh=hg}$, which is a special result by Burnside’s theorem.
I found that the number of conjugacy class in $D_8$ is 5, so to double check, I listed down all $C_{G}(g)={h in G ; gh=hg}$.
Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.
Well, $|C_G(1)|=|D_8|=16$,
$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7}|=8$,
$|C_G(r^4)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m}|=9$ since $r^4m=mr^4$
Similarly, we can count for the reflections;
$|C_G(m)|=|C_G(r^4m)|=|{1,m}|=2$, while the rest of the elements with any reflections only has one element, i.e $|C_G(r^nm)|=|{1}|=1,n neq 0,4$
So the problem comes that my summation is 83, while my $|G|=16$. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...
abstract-algebra group-theory finite-groups group-actions dihedral-groups
edited Nov 27 '18 at 4:17
Shubham
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
I’m using the formula that the number of conjugacy class is given to be $frac{1}{|G|}sum|C_{G}(g)|$, where $C_{G}(g)={h in G ; gh=hg}$, which is a special result by Burnside’s theorem.
I found that the number of conjugacy class in $D_8$ is 5, so to double check, I listed down all $C_{G}(g)={h in G ; gh=hg}$.
Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.
Well, $|C_G(1)|=|D_8|=16$,
$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7}|=8$,
$|C_G(r^4)|=|{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m}|=9$ since $r^4m=mr^4$
Similarly, we can count for the reflections;
$|C_G(m)|=|C_G(r^4m)|=|{1,m}|=2$, while the rest of the elements with any reflections only has one element, i.e $|C_G(r^nm)|=|{1}|=1,n neq 0,4$
So the problem comes that my summation is 83, while my $|G|=16$. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...
abstract-algebra group-theory finite-groups group-actions dihedral-groups
abstract-algebra group-theory finite-groups group-actions dihedral-groups
edited Nov 27 '18 at 4:17
Shubham
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
edited Nov 27 '18 at 4:17
Shubham
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
edited Nov 27 '18 at 4:17
Shubham
1,5851519
edited Nov 27 '18 at 4:17
Shubham
1,5851519
edited Nov 27 '18 at 4:17
Shubham
1,5851519
1,5851519
asked Nov 27 '18 at 0:22
Icycarus
4721213
asked Nov 27 '18 at 0:22
Icycarus
4721213
asked Nov 27 '18 at 0:22
Icycarus
4721213
4721213
The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10
add a comment |
The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10
The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10
add a comment |
2 Answers
2
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votes
$G = {r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
${e}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
${r, r^7},{r^2, r^6},{r^3, r^5}, {r^4}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of ${ar, ar^3, ar^5, ar^7}$ and ${a, ar^2, ar^4, ar^6}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\
|C_G(r)| = 8 text { times 6}\
|C_G(r^4)| = 16\
|C_G(a)| = 4 text { times 8}$
$frac {16times 2 + 8times 6 + 4times 8}{16} = 7$
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
add a comment |
$D_n$
In general its class equation given by following
Case 1: n is odd
$Z(D_n)=${$e$}
$(n-1)/2$ classes of {$r^i,r^-i$} for $1leq ileq (n-1)/2$
$1$ class of {$sr^i|0leq i leq n-1$}
SO class equation $n=1+2.(n-1)/2+n$
Case 2:n is even
$Z(D_n)=${$e,r^{n/2}$}
$(n)/2-1$ classes of {$r^i,r^-i$} for $1leq ileq (n)/2-1$
$1$ class of {$sr^{2i}|0leq i leq n/2-1$}
$1$ class of {$sr^{2i+1}|0leq i leq n/2-1$}
SO class equation $n=1+1+2.(n/2-1)+n/2.1+n/2.1$
In specific:
$D_8$
$1+1+2+2+2+4+4=16 $is class equation
answered Nov 27 '18 at 3:49
Shubham
1,5851519
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
add a comment |
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2 Answers
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2 Answers
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oldest
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active
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$G = {r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
${e}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
${r, r^7},{r^2, r^6},{r^3, r^5}, {r^4}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of ${ar, ar^3, ar^5, ar^7}$ and ${a, ar^2, ar^4, ar^6}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\
|C_G(r)| = 8 text { times 6}\
|C_G(r^4)| = 16\
|C_G(a)| = 4 text { times 8}$
$frac {16times 2 + 8times 6 + 4times 8}{16} = 7$
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
add a comment |
$G = {r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
${e}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
${r, r^7},{r^2, r^6},{r^3, r^5}, {r^4}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of ${ar, ar^3, ar^5, ar^7}$ and ${a, ar^2, ar^4, ar^6}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\
|C_G(r)| = 8 text { times 6}\
|C_G(r^4)| = 16\
|C_G(a)| = 4 text { times 8}$
$frac {16times 2 + 8times 6 + 4times 8}{16} = 7$
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
add a comment |
$G = {r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
${e}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
${r, r^7},{r^2, r^6},{r^3, r^5}, {r^4}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of ${ar, ar^3, ar^5, ar^7}$ and ${a, ar^2, ar^4, ar^6}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\
|C_G(r)| = 8 text { times 6}\
|C_G(r^4)| = 16\
|C_G(a)| = 4 text { times 8}$
$frac {16times 2 + 8times 6 + 4times 8}{16} = 7$
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
$G = {r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
${e}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
${r, r^7},{r^2, r^6},{r^3, r^5}, {r^4}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of ${ar, ar^3, ar^5, ar^7}$ and ${a, ar^2, ar^4, ar^6}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\
|C_G(r)| = 8 text { times 6}\
|C_G(r^4)| = 16\
|C_G(a)| = 4 text { times 8}$
$frac {16times 2 + 8times 6 + 4times 8}{16} = 7$
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
edited Nov 27 '18 at 1:56
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
answered Nov 27 '18 at 1:13
Doug M
43.9k31854
43.9k31854
add a comment |
add a comment |
$D_n$
In general its class equation given by following
Case 1: n is odd
$Z(D_n)=${$e$}
$(n-1)/2$ classes of {$r^i,r^-i$} for $1leq ileq (n-1)/2$
$1$ class of {$sr^i|0leq i leq n-1$}
SO class equation $n=1+2.(n-1)/2+n$
Case 2:n is even
$Z(D_n)=${$e,r^{n/2}$}
$(n)/2-1$ classes of {$r^i,r^-i$} for $1leq ileq (n)/2-1$
$1$ class of {$sr^{2i}|0leq i leq n/2-1$}
$1$ class of {$sr^{2i+1}|0leq i leq n/2-1$}
SO class equation $n=1+1+2.(n/2-1)+n/2.1+n/2.1$
In specific:
$D_8$
$1+1+2+2+2+4+4=16 $is class equation
answered Nov 27 '18 at 3:49
Shubham
1,5851519
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
add a comment |
$D_n$
In general its class equation given by following
Case 1: n is odd
$Z(D_n)=${$e$}
$(n-1)/2$ classes of {$r^i,r^-i$} for $1leq ileq (n-1)/2$
$1$ class of {$sr^i|0leq i leq n-1$}
SO class equation $n=1+2.(n-1)/2+n$
Case 2:n is even
$Z(D_n)=${$e,r^{n/2}$}
$(n)/2-1$ classes of {$r^i,r^-i$} for $1leq ileq (n)/2-1$
$1$ class of {$sr^{2i}|0leq i leq n/2-1$}
$1$ class of {$sr^{2i+1}|0leq i leq n/2-1$}
SO class equation $n=1+1+2.(n/2-1)+n/2.1+n/2.1$
In specific:
$D_8$
$1+1+2+2+2+4+4=16 $is class equation
answered Nov 27 '18 at 3:49
Shubham
1,5851519
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
add a comment |
$D_n$
In general its class equation given by following
Case 1: n is odd
$Z(D_n)=${$e$}
$(n-1)/2$ classes of {$r^i,r^-i$} for $1leq ileq (n-1)/2$
$1$ class of {$sr^i|0leq i leq n-1$}
SO class equation $n=1+2.(n-1)/2+n$
Case 2:n is even
$Z(D_n)=${$e,r^{n/2}$}
$(n)/2-1$ classes of {$r^i,r^-i$} for $1leq ileq (n)/2-1$
$1$ class of {$sr^{2i}|0leq i leq n/2-1$}
$1$ class of {$sr^{2i+1}|0leq i leq n/2-1$}
SO class equation $n=1+1+2.(n/2-1)+n/2.1+n/2.1$
In specific:
$D_8$
$1+1+2+2+2+4+4=16 $is class equation
answered Nov 27 '18 at 3:49
Shubham
1,5851519
$D_n$
In general its class equation given by following
Case 1: n is odd
$Z(D_n)=${$e$}
$(n-1)/2$ classes of {$r^i,r^-i$} for $1leq ileq (n-1)/2$
$1$ class of {$sr^i|0leq i leq n-1$}
SO class equation $n=1+2.(n-1)/2+n$
Case 2:n is even
$Z(D_n)=${$e,r^{n/2}$}
$(n)/2-1$ classes of {$r^i,r^-i$} for $1leq ileq (n)/2-1$
$1$ class of {$sr^{2i}|0leq i leq n/2-1$}
$1$ class of {$sr^{2i+1}|0leq i leq n/2-1$}
SO class equation $n=1+1+2.(n/2-1)+n/2.1+n/2.1$
In specific:
$D_8$
$1+1+2+2+2+4+4=16 $is class equation
answered Nov 27 '18 at 3:49
Shubham
1,5851519
edited Nov 27 '18 at 16:06
answered Nov 27 '18 at 3:49
Shubham
1,5851519
answered Nov 27 '18 at 3:49
Shubham
1,5851519
answered Nov 27 '18 at 3:49
Shubham
1,5851519
1,5851519
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
add a comment |
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
In that case, D8 has 16 class equations from your answer(?) Is that right?
– Icycarus
Nov 27 '18 at 16:00
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
No . There are 7 classes I have mentioned now
– Shubham
Nov 27 '18 at 16:07
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Ah okay! I get what you mean now! There are 3 classes with 2 elements and 2 classes with 4 elements! Right, thanks so much!
– Icycarus
Nov 27 '18 at 16:08
Its Okay........
– Shubham
Nov 27 '18 at 16:09
Its Okay........
– Shubham
Nov 27 '18 at 16:09
add a comment |
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The centralizer is a subgroup. A subgroup of a group of order 16 can't have order 9 (the order of a subgroup divides the order of the group).
– Gerry Myerson
Nov 27 '18 at 3:21
Also, every non-identity element commutes with at least itself and the identity, so has centralizer of size at least 2.
– Gerry Myerson
Nov 27 '18 at 3:23
Thanks for pointing it out! I will look into it and edit it as well! I might have looked at the wrong source telling me that D8 has only 5 conjugacy classes
– Icycarus
Nov 27 '18 at 15:58
Some authors use the notation $D_8$ for the 8-element dihedral group. That group has only five conjugacy classes.
– Gerry Myerson
Nov 27 '18 at 21:10