Character Table Dihedral group of $D_6$
$begingroup$
I'm having real troubles with finding the character table of the dihedral group $D_6$ of order 12: $D_6 = langle a,b |a^6 = 1 , b^2 = 1, aba = b rangle$. I've already found the conjugacy classes: ${1}, {a, a^5} , {a^2,a^4 }, {a^3} , {b,ba^2,ba^4 }$ and ${ba,ba^3,ba^5}$. But from there, I'm completely stuck.
Any help would be dearly appreciated.
group-theory representation-theory characters dihedral-groups
asked Dec 31 '14 at 10:46
RileyRiley
643414
$endgroup$
add a comment |
$begingroup$
I'm having real troubles with finding the character table of the dihedral group $D_6$ of order 12: $D_6 = langle a,b |a^6 = 1 , b^2 = 1, aba = b rangle$. I've already found the conjugacy classes: ${1}, {a, a^5} , {a^2,a^4 }, {a^3} , {b,ba^2,ba^4 }$ and ${ba,ba^3,ba^5}$. But from there, I'm completely stuck.
Any help would be dearly appreciated.
group-theory representation-theory characters dihedral-groups
asked Dec 31 '14 at 10:46
RileyRiley
643414
$endgroup$
2
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27
add a comment |
$begingroup$
I'm having real troubles with finding the character table of the dihedral group $D_6$ of order 12: $D_6 = langle a,b |a^6 = 1 , b^2 = 1, aba = b rangle$. I've already found the conjugacy classes: ${1}, {a, a^5} , {a^2,a^4 }, {a^3} , {b,ba^2,ba^4 }$ and ${ba,ba^3,ba^5}$. But from there, I'm completely stuck.
Any help would be dearly appreciated.
group-theory representation-theory characters dihedral-groups
asked Dec 31 '14 at 10:46
RileyRiley
643414
$endgroup$
I'm having real troubles with finding the character table of the dihedral group $D_6$ of order 12: $D_6 = langle a,b |a^6 = 1 , b^2 = 1, aba = b rangle$. I've already found the conjugacy classes: ${1}, {a, a^5} , {a^2,a^4 }, {a^3} , {b,ba^2,ba^4 }$ and ${ba,ba^3,ba^5}$. But from there, I'm completely stuck.
Any help would be dearly appreciated.
group-theory representation-theory characters dihedral-groups
group-theory representation-theory characters dihedral-groups
asked Dec 31 '14 at 10:46
RileyRiley
643414
asked Dec 31 '14 at 10:46
RileyRiley
643414
edited Jan 23 '15 at 17:32
Riley
asked Dec 31 '14 at 10:46
RileyRiley
643414
asked Dec 31 '14 at 10:46
RileyRiley
643414
asked Dec 31 '14 at 10:46
RileyRiley
643414
643414
2
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27
add a comment |
2
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27
2
2
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Found it! For the people who need an extra hand, here's a sketch of how to do it:
First, determine the conjugacy classes. Amount conjugacy classes= amount of irreducible representations.
Second, determine these representations. The one-dimensional representations can be found via the normal subgroup properties. You'll find that there are four one-dimensional and two two-dimensional irreducible representations, you can find those by the product formula for 2 representations.
answered Jan 23 '15 at 11:23
RileyRiley
643414
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Found it! For the people who need an extra hand, here's a sketch of how to do it:
First, determine the conjugacy classes. Amount conjugacy classes= amount of irreducible representations.
Second, determine these representations. The one-dimensional representations can be found via the normal subgroup properties. You'll find that there are four one-dimensional and two two-dimensional irreducible representations, you can find those by the product formula for 2 representations.
answered Jan 23 '15 at 11:23
RileyRiley
643414
$endgroup$
add a comment |
$begingroup$
Found it! For the people who need an extra hand, here's a sketch of how to do it:
First, determine the conjugacy classes. Amount conjugacy classes= amount of irreducible representations.
Second, determine these representations. The one-dimensional representations can be found via the normal subgroup properties. You'll find that there are four one-dimensional and two two-dimensional irreducible representations, you can find those by the product formula for 2 representations.
answered Jan 23 '15 at 11:23
RileyRiley
643414
$endgroup$
add a comment |
$begingroup$
Found it! For the people who need an extra hand, here's a sketch of how to do it:
First, determine the conjugacy classes. Amount conjugacy classes= amount of irreducible representations.
Second, determine these representations. The one-dimensional representations can be found via the normal subgroup properties. You'll find that there are four one-dimensional and two two-dimensional irreducible representations, you can find those by the product formula for 2 representations.
answered Jan 23 '15 at 11:23
RileyRiley
643414
$endgroup$
Found it! For the people who need an extra hand, here's a sketch of how to do it:
First, determine the conjugacy classes. Amount conjugacy classes= amount of irreducible representations.
Second, determine these representations. The one-dimensional representations can be found via the normal subgroup properties. You'll find that there are four one-dimensional and two two-dimensional irreducible representations, you can find those by the product formula for 2 representations.
answered Jan 23 '15 at 11:23
RileyRiley
643414
answered Jan 23 '15 at 11:23
RileyRiley
643414
answered Jan 23 '15 at 11:23
RileyRiley
643414
answered Jan 23 '15 at 11:23
RileyRiley
643414
643414
add a comment |
add a comment |
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2
$begingroup$
Start by finding the linear characters by finding the derived subgroup.
$endgroup$
– Tobias Kildetoft
Dec 31 '14 at 11:27