Wedderburn Decomposition by using Clifford theorem.
$begingroup$
Let $H$ be a normal subgroup of $G$ and we know Wedderburn decomposition of semi simple algebra $FH$ over a finite field $F$ as $$FH=Foplus M_{3}(F)oplus M_{3}(F)oplus M_{4}(F)oplus M_{5}(F).$$ But i want Wedderburn decomposition of $FG$. After my all calculation i found that $$FG=Foplus M_{n_1}(F)oplus M_{n_2}(F)oplus M_{n_3}(F)oplus M_{n_4}(F)$$ and two choices of $n_{i}$ as $2,3,4,5$ and $4,4,5,5,,6.$ Now i am confused which one is answer. I thought Clifford theorem will help but i am unable to apply it. Please help me apply to it or any other idea. Thanks.
modules representation-theory group-rings
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
$endgroup$
|
show 3 more comments
$begingroup$
Let $H$ be a normal subgroup of $G$ and we know Wedderburn decomposition of semi simple algebra $FH$ over a finite field $F$ as $$FH=Foplus M_{3}(F)oplus M_{3}(F)oplus M_{4}(F)oplus M_{5}(F).$$ But i want Wedderburn decomposition of $FG$. After my all calculation i found that $$FG=Foplus M_{n_1}(F)oplus M_{n_2}(F)oplus M_{n_3}(F)oplus M_{n_4}(F)$$ and two choices of $n_{i}$ as $2,3,4,5$ and $4,4,5,5,,6.$ Now i am confused which one is answer. I thought Clifford theorem will help but i am unable to apply it. Please help me apply to it or any other idea. Thanks.
modules representation-theory group-rings
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
$endgroup$
$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04
|
show 3 more comments
$begingroup$
Let $H$ be a normal subgroup of $G$ and we know Wedderburn decomposition of semi simple algebra $FH$ over a finite field $F$ as $$FH=Foplus M_{3}(F)oplus M_{3}(F)oplus M_{4}(F)oplus M_{5}(F).$$ But i want Wedderburn decomposition of $FG$. After my all calculation i found that $$FG=Foplus M_{n_1}(F)oplus M_{n_2}(F)oplus M_{n_3}(F)oplus M_{n_4}(F)$$ and two choices of $n_{i}$ as $2,3,4,5$ and $4,4,5,5,,6.$ Now i am confused which one is answer. I thought Clifford theorem will help but i am unable to apply it. Please help me apply to it or any other idea. Thanks.
modules representation-theory group-rings
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
$endgroup$
Let $H$ be a normal subgroup of $G$ and we know Wedderburn decomposition of semi simple algebra $FH$ over a finite field $F$ as $$FH=Foplus M_{3}(F)oplus M_{3}(F)oplus M_{4}(F)oplus M_{5}(F).$$ But i want Wedderburn decomposition of $FG$. After my all calculation i found that $$FG=Foplus M_{n_1}(F)oplus M_{n_2}(F)oplus M_{n_3}(F)oplus M_{n_4}(F)$$ and two choices of $n_{i}$ as $2,3,4,5$ and $4,4,5,5,,6.$ Now i am confused which one is answer. I thought Clifford theorem will help but i am unable to apply it. Please help me apply to it or any other idea. Thanks.
modules representation-theory group-rings
modules representation-theory group-rings
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
edited Dec 28 '18 at 7:39
Jyrki Lahtonen
109k13169372
109k13169372
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
asked Dec 26 '18 at 2:30
neelkanthneelkanth
2,15121028
2,15121028
$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04
|
show 3 more comments
$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04
$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04
|
show 3 more comments
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$begingroup$
Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what?
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:34
$begingroup$
Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:37
$begingroup$
Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:41
$begingroup$
@JyrkiLahtonen yes sir decomposition of normal subgroup is given...
$endgroup$
– neelkanth
Dec 28 '18 at 15:02
$begingroup$
Sir H is of index $2$ i.e. $G$ is of order $120.$
$endgroup$
– neelkanth
Dec 28 '18 at 15:04