Find all irreducible characters of a matrix group on finite field $mathbb F_5$
$begingroup$
Find all irreducible characters of matrix group $G =left{ left( begin{array}{cc}
a & b \0 & a^{-1}end{array} right)|,,, a,b inmathbb F_5, anot=0 right}$.
The former question is to find irreducible characters of subgroup of $G$
$H =left{ left( begin{array}{cc}
a & 0 \0 & a^{-1}end{array} right)|, ainmathbb F_5^{times} right}$ and
$N =left{ left( begin{array}{cc}
1 & b \0 & 1 end{array} right)|, binmathbb F_5 right}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
abstract-algebra group-theory representation-theory characters
$endgroup$
add a comment |
$begingroup$
Find all irreducible characters of matrix group $G =left{ left( begin{array}{cc}
a & b \0 & a^{-1}end{array} right)|,,, a,b inmathbb F_5, anot=0 right}$.
The former question is to find irreducible characters of subgroup of $G$
$H =left{ left( begin{array}{cc}
a & 0 \0 & a^{-1}end{array} right)|, ainmathbb F_5^{times} right}$ and
$N =left{ left( begin{array}{cc}
1 & b \0 & 1 end{array} right)|, binmathbb F_5 right}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
abstract-algebra group-theory representation-theory characters
$endgroup$
$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23
add a comment |
$begingroup$
Find all irreducible characters of matrix group $G =left{ left( begin{array}{cc}
a & b \0 & a^{-1}end{array} right)|,,, a,b inmathbb F_5, anot=0 right}$.
The former question is to find irreducible characters of subgroup of $G$
$H =left{ left( begin{array}{cc}
a & 0 \0 & a^{-1}end{array} right)|, ainmathbb F_5^{times} right}$ and
$N =left{ left( begin{array}{cc}
1 & b \0 & 1 end{array} right)|, binmathbb F_5 right}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
abstract-algebra group-theory representation-theory characters
$endgroup$
Find all irreducible characters of matrix group $G =left{ left( begin{array}{cc}
a & b \0 & a^{-1}end{array} right)|,,, a,b inmathbb F_5, anot=0 right}$.
The former question is to find irreducible characters of subgroup of $G$
$H =left{ left( begin{array}{cc}
a & 0 \0 & a^{-1}end{array} right)|, ainmathbb F_5^{times} right}$ and
$N =left{ left( begin{array}{cc}
1 & b \0 & 1 end{array} right)|, binmathbb F_5 right}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
abstract-algebra group-theory representation-theory characters
abstract-algebra group-theory representation-theory characters
edited Jan 6 at 22:57
Andrews
asked Jan 1 at 15:45
AndrewsAndrews
7361318
7361318
$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23
add a comment |
$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23
$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $mathbb{C}$. As indicated $G=HN$, $N lhd G$, $H cap N=1$ with $H cong C_4$ and $N cong C_5$. So $|G|=20$. Note that $G/N cong C_4$, whence abelian, so $G' subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)={pm I}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $chi in Irr(G)$ is non-linear, then $chi(1) leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
$endgroup$
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
add a comment |
$begingroup$
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $chi_1,dotschi_4$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm4A&rm4Bcr
rm size&1&1&1&1cr
hline
rho_{1}&1&1&1&1cr
rho_{2}&1&1&-1&-1cr
rho_{3}&1&-1&-i&icr
rho_{4}&1&-1&i&-icr
end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $chi_4,dots,chi_8$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm5A&rm5Bcr
rm size&1&5&2&2cr
hline
eta_{1}&1&1&1&1cr
eta_{2}&1&-1&1&1cr
eta_{3}&2&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
eta_{4}&2&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
begin{array}{c|rrrrrrrr}
rm class&rm I&rm -I&rm4A&rm4B&rm5A&rm5B&rm10A&rm10Bcr
rm size&1&1&5&5&2&2&2&2cr
hline
chi_{1}&1&1&1&1&1&1&1&1cr
chi_{2}&1&1&-1&-1&1&1&1&1cr
chi_{3}&1&-1&-i&i&1&1&-1&-1cr
chi_{4}&1&-1&i&-i&1&1&-1&-1cr
chi_{5}&2&2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
chi_{6}&2&2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
chi_{7}&2&-2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
chi_{8}&2&-2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
end{array}
$$
$endgroup$
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $mathbb{C}$. As indicated $G=HN$, $N lhd G$, $H cap N=1$ with $H cong C_4$ and $N cong C_5$. So $|G|=20$. Note that $G/N cong C_4$, whence abelian, so $G' subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)={pm I}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $chi in Irr(G)$ is non-linear, then $chi(1) leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
$endgroup$
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
add a comment |
$begingroup$
(Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $mathbb{C}$. As indicated $G=HN$, $N lhd G$, $H cap N=1$ with $H cong C_4$ and $N cong C_5$. So $|G|=20$. Note that $G/N cong C_4$, whence abelian, so $G' subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)={pm I}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $chi in Irr(G)$ is non-linear, then $chi(1) leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
$endgroup$
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
add a comment |
$begingroup$
(Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $mathbb{C}$. As indicated $G=HN$, $N lhd G$, $H cap N=1$ with $H cong C_4$ and $N cong C_5$. So $|G|=20$. Note that $G/N cong C_4$, whence abelian, so $G' subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)={pm I}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $chi in Irr(G)$ is non-linear, then $chi(1) leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
$endgroup$
(Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $mathbb{C}$. As indicated $G=HN$, $N lhd G$, $H cap N=1$ with $H cong C_4$ and $N cong C_5$. So $|G|=20$. Note that $G/N cong C_4$, whence abelian, so $G' subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)={pm I}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $chi in Irr(G)$ is non-linear, then $chi(1) leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
answered Jan 2 at 9:59
Nicky HeksterNicky Hekster
28.8k63456
28.8k63456
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
add a comment |
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
2
2
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters.
$endgroup$
– Derek Holt
Jan 2 at 10:10
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
$begingroup$
Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 times 8$ character table.
$endgroup$
– Nicky Hekster
Jan 2 at 10:41
add a comment |
$begingroup$
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $chi_1,dotschi_4$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm4A&rm4Bcr
rm size&1&1&1&1cr
hline
rho_{1}&1&1&1&1cr
rho_{2}&1&1&-1&-1cr
rho_{3}&1&-1&-i&icr
rho_{4}&1&-1&i&-icr
end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $chi_4,dots,chi_8$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm5A&rm5Bcr
rm size&1&5&2&2cr
hline
eta_{1}&1&1&1&1cr
eta_{2}&1&-1&1&1cr
eta_{3}&2&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
eta_{4}&2&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
begin{array}{c|rrrrrrrr}
rm class&rm I&rm -I&rm4A&rm4B&rm5A&rm5B&rm10A&rm10Bcr
rm size&1&1&5&5&2&2&2&2cr
hline
chi_{1}&1&1&1&1&1&1&1&1cr
chi_{2}&1&1&-1&-1&1&1&1&1cr
chi_{3}&1&-1&-i&i&1&1&-1&-1cr
chi_{4}&1&-1&i&-i&1&1&-1&-1cr
chi_{5}&2&2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
chi_{6}&2&2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
chi_{7}&2&-2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
chi_{8}&2&-2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
end{array}
$$
$endgroup$
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
add a comment |
$begingroup$
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $chi_1,dotschi_4$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm4A&rm4Bcr
rm size&1&1&1&1cr
hline
rho_{1}&1&1&1&1cr
rho_{2}&1&1&-1&-1cr
rho_{3}&1&-1&-i&icr
rho_{4}&1&-1&i&-icr
end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $chi_4,dots,chi_8$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm5A&rm5Bcr
rm size&1&5&2&2cr
hline
eta_{1}&1&1&1&1cr
eta_{2}&1&-1&1&1cr
eta_{3}&2&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
eta_{4}&2&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
begin{array}{c|rrrrrrrr}
rm class&rm I&rm -I&rm4A&rm4B&rm5A&rm5B&rm10A&rm10Bcr
rm size&1&1&5&5&2&2&2&2cr
hline
chi_{1}&1&1&1&1&1&1&1&1cr
chi_{2}&1&1&-1&-1&1&1&1&1cr
chi_{3}&1&-1&-i&i&1&1&-1&-1cr
chi_{4}&1&-1&i&-i&1&1&-1&-1cr
chi_{5}&2&2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
chi_{6}&2&2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
chi_{7}&2&-2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
chi_{8}&2&-2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
end{array}
$$
$endgroup$
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
add a comment |
$begingroup$
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $chi_1,dotschi_4$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm4A&rm4Bcr
rm size&1&1&1&1cr
hline
rho_{1}&1&1&1&1cr
rho_{2}&1&1&-1&-1cr
rho_{3}&1&-1&-i&icr
rho_{4}&1&-1&i&-icr
end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $chi_4,dots,chi_8$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm5A&rm5Bcr
rm size&1&5&2&2cr
hline
eta_{1}&1&1&1&1cr
eta_{2}&1&-1&1&1cr
eta_{3}&2&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
eta_{4}&2&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
begin{array}{c|rrrrrrrr}
rm class&rm I&rm -I&rm4A&rm4B&rm5A&rm5B&rm10A&rm10Bcr
rm size&1&1&5&5&2&2&2&2cr
hline
chi_{1}&1&1&1&1&1&1&1&1cr
chi_{2}&1&1&-1&-1&1&1&1&1cr
chi_{3}&1&-1&-i&i&1&1&-1&-1cr
chi_{4}&1&-1&i&-i&1&1&-1&-1cr
chi_{5}&2&2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
chi_{6}&2&2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
chi_{7}&2&-2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
chi_{8}&2&-2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
end{array}
$$
$endgroup$
From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$. Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution. This combined with the character table of $C_4$ (below) gives us $chi_1,dotschi_4$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm4A&rm4Bcr
rm size&1&1&1&1cr
hline
rho_{1}&1&1&1&1cr
rho_{2}&1&1&-1&-1cr
rho_{3}&1&-1&-i&icr
rho_{4}&1&-1&i&-icr
end{array}
$$
All that remains now is to find a 2-dimensional representation. We can do this by calculating the quotient group $G/Z(G)$. It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go). Now, we can lift a two dimensional character from $D_5$ (below), say $eta_3$ and then multiply it by the three linear characters we have already calculated. This will give us $chi_4,dots,chi_8$.
$$
begin{array}{c|rrrr}
rm class&rm1&rm2&rm5A&rm5Bcr
rm size&1&5&2&2cr
hline
eta_{1}&1&1&1&1cr
eta_{2}&1&-1&1&1cr
eta_{3}&2&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
eta_{4}&2&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly. The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
begin{array}{c|rrrrrrrr}
rm class&rm I&rm -I&rm4A&rm4B&rm5A&rm5B&rm10A&rm10Bcr
rm size&1&1&5&5&2&2&2&2cr
hline
chi_{1}&1&1&1&1&1&1&1&1cr
chi_{2}&1&1&-1&-1&1&1&1&1cr
chi_{3}&1&-1&-i&i&1&1&-1&-1cr
chi_{4}&1&-1&i&-i&1&1&-1&-1cr
chi_{5}&2&2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}cr
chi_{6}&2&2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}cr
chi_{7}&2&-2&0&0&frac{-1+sqrt{5}}{2}&frac{-1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
chi_{8}&2&-2&0&0&frac{-1-sqrt{5}}{2}&frac{-1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
end{array}
$$
answered Jan 8 at 17:49
Sam HughesSam Hughes
552112
552112
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
add a comment |
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
1
1
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
$begingroup$
+1 from me! I returned to the question today and see that you already have done a great job!
$endgroup$
– Nicky Hekster
Jan 8 at 18:14
add a comment |
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$begingroup$
You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$.
$endgroup$
– Derek Holt
Jan 1 at 16:03
$begingroup$
Derek says that restricting to $a = pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $psi : A to mathbb{C}^times$ then $Ind_A^G psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$
$endgroup$
– reuns
Jan 1 at 16:50
$begingroup$
You asked for a hint, not a detailed answer.
$endgroup$
– Derek Holt
Jan 6 at 18:23