Why is a unique Sylow p-subgroup normal?
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I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.
abstract-algebra normal-subgroups sylow-theory
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show 2 more comments
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I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.
abstract-algebra normal-subgroups sylow-theory
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I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
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– Foobanana
Nov 2 '17 at 16:14
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$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
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– Randall
Nov 2 '17 at 16:14
4
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You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
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– Bungo
Nov 2 '17 at 16:15
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I think @Bungo put it better than I did :)
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– Foobanana
Nov 2 '17 at 16:16
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@Foobanana Your remark is quite correct, I just put a bit more detail :-)
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– Bungo
Nov 2 '17 at 16:17
|
show 2 more comments
$begingroup$
I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.
abstract-algebra normal-subgroups sylow-theory
$endgroup$
I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.
abstract-algebra normal-subgroups sylow-theory
abstract-algebra normal-subgroups sylow-theory
edited Nov 2 '17 at 16:28
Teddy38
2,1912520
2,1912520
asked Nov 2 '17 at 16:12
mikkelmkmikkelmk
695
695
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I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14
$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14
4
$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15
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I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16
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@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17
|
show 2 more comments
$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14
$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14
4
$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15
$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16
$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17
$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14
$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14
$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14
$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14
4
4
$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15
$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15
$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16
$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16
$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17
$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17
|
show 2 more comments
1 Answer
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$begingroup$
Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.
$endgroup$
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$begingroup$
Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.
$endgroup$
add a comment |
$begingroup$
Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.
$endgroup$
add a comment |
$begingroup$
Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.
$endgroup$
Suppose that $H$ is the unique $p$-Sylow subgroup, for any $gin G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $hin H, ghg^{-1}in H$ and $H$ is normal.
answered Nov 2 '17 at 16:14
Tsemo AristideTsemo Aristide
57.9k11445
57.9k11445
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$begingroup$
I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:14
$begingroup$
$N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what?
$endgroup$
– Randall
Nov 2 '17 at 16:14
4
$begingroup$
You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal.
$endgroup$
– Bungo
Nov 2 '17 at 16:15
$begingroup$
I think @Bungo put it better than I did :)
$endgroup$
– Foobanana
Nov 2 '17 at 16:16
$begingroup$
@Foobanana Your remark is quite correct, I just put a bit more detail :-)
$endgroup$
– Bungo
Nov 2 '17 at 16:17