Action of group of order $(p-1)p$ on a $mathbb F_p$ vector space
up vote
1
down vote
favorite
By classification, I find that any group of order $42$ cannot act on a finite dimensional $mathbb F_7$ vector space $V$ such the representation is irreducible and faithful. Does it hold generally? Assume a group of order $(p-1)p$ on a $mathbb F_p$ vector space $V$, could the representation be both irreducible and faithful?
abstract-algebra group-theory
asked Nov 23 at 12:17
zzy
2,2281419
add a comment |
up vote
1
down vote
favorite
By classification, I find that any group of order $42$ cannot act on a finite dimensional $mathbb F_7$ vector space $V$ such the representation is irreducible and faithful. Does it hold generally? Assume a group of order $(p-1)p$ on a $mathbb F_p$ vector space $V$, could the representation be both irreducible and faithful?
abstract-algebra group-theory
asked Nov 23 at 12:17
zzy
2,2281419
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
By classification, I find that any group of order $42$ cannot act on a finite dimensional $mathbb F_7$ vector space $V$ such the representation is irreducible and faithful. Does it hold generally? Assume a group of order $(p-1)p$ on a $mathbb F_p$ vector space $V$, could the representation be both irreducible and faithful?
abstract-algebra group-theory
asked Nov 23 at 12:17
zzy
2,2281419
By classification, I find that any group of order $42$ cannot act on a finite dimensional $mathbb F_7$ vector space $V$ such the representation is irreducible and faithful. Does it hold generally? Assume a group of order $(p-1)p$ on a $mathbb F_p$ vector space $V$, could the representation be both irreducible and faithful?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 23 at 12:17
zzy
2,2281419
asked Nov 23 at 12:17
zzy
2,2281419
asked Nov 23 at 12:17
zzy
2,2281419
asked Nov 23 at 12:17
zzy
2,2281419
asked Nov 23 at 12:17
zzy
2,2281419
2,2281419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Any group of order $(p-1)p$ has a normal subgroup of order $p$ by Sylow's Theorem.
In general, if $N$ is a normal $p$-subgroup of any finite group $G$, then $N$ is in the kernel of any irreducible representation of $G$ over a field with characteristic $p$. (This is because the only irreducible represenation of $N$ in characteristic $p$ is the trivial one, so $N$ must have fixed points in its action on the $G$-module, and hence it must act trivially by irreducibility of the $G$-module.) So if $1 ne N$, then $G$ has no faithful irreducible representation over a field with characteristic $p$.
So the answer to your question is no.
answered Nov 23 at 14:11
Derek Holt
52.2k53570
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Any group of order $(p-1)p$ has a normal subgroup of order $p$ by Sylow's Theorem.
In general, if $N$ is a normal $p$-subgroup of any finite group $G$, then $N$ is in the kernel of any irreducible representation of $G$ over a field with characteristic $p$. (This is because the only irreducible represenation of $N$ in characteristic $p$ is the trivial one, so $N$ must have fixed points in its action on the $G$-module, and hence it must act trivially by irreducibility of the $G$-module.) So if $1 ne N$, then $G$ has no faithful irreducible representation over a field with characteristic $p$.
So the answer to your question is no.
answered Nov 23 at 14:11
Derek Holt
52.2k53570
add a comment |
up vote
4
down vote
accepted
Any group of order $(p-1)p$ has a normal subgroup of order $p$ by Sylow's Theorem.
In general, if $N$ is a normal $p$-subgroup of any finite group $G$, then $N$ is in the kernel of any irreducible representation of $G$ over a field with characteristic $p$. (This is because the only irreducible represenation of $N$ in characteristic $p$ is the trivial one, so $N$ must have fixed points in its action on the $G$-module, and hence it must act trivially by irreducibility of the $G$-module.) So if $1 ne N$, then $G$ has no faithful irreducible representation over a field with characteristic $p$.
So the answer to your question is no.
answered Nov 23 at 14:11
Derek Holt
52.2k53570
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Any group of order $(p-1)p$ has a normal subgroup of order $p$ by Sylow's Theorem.
In general, if $N$ is a normal $p$-subgroup of any finite group $G$, then $N$ is in the kernel of any irreducible representation of $G$ over a field with characteristic $p$. (This is because the only irreducible represenation of $N$ in characteristic $p$ is the trivial one, so $N$ must have fixed points in its action on the $G$-module, and hence it must act trivially by irreducibility of the $G$-module.) So if $1 ne N$, then $G$ has no faithful irreducible representation over a field with characteristic $p$.
So the answer to your question is no.
answered Nov 23 at 14:11
Derek Holt
52.2k53570
Any group of order $(p-1)p$ has a normal subgroup of order $p$ by Sylow's Theorem.
In general, if $N$ is a normal $p$-subgroup of any finite group $G$, then $N$ is in the kernel of any irreducible representation of $G$ over a field with characteristic $p$. (This is because the only irreducible represenation of $N$ in characteristic $p$ is the trivial one, so $N$ must have fixed points in its action on the $G$-module, and hence it must act trivially by irreducibility of the $G$-module.) So if $1 ne N$, then $G$ has no faithful irreducible representation over a field with characteristic $p$.
So the answer to your question is no.
answered Nov 23 at 14:11
Derek Holt
52.2k53570
answered Nov 23 at 14:11
Derek Holt
52.2k53570
answered Nov 23 at 14:11
Derek Holt
52.2k53570
answered Nov 23 at 14:11
Derek Holt
52.2k53570
52.2k53570
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010293%2faction-of-group-of-order-p-1p-on-a-mathbb-f-p-vector-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
