In a group of order 400, is the normalizer of one of the 16 Sylow 5-subgroups itself?
$begingroup$
In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?
I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!
abstract-algebra group-theory sylow-theory
asked Jan 6 at 22:06
pendermathpendermath
52910
$endgroup$
add a comment |
$begingroup$
In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?
I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!
abstract-algebra group-theory sylow-theory
asked Jan 6 at 22:06
pendermathpendermath
52910
$endgroup$
add a comment |
$begingroup$
In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?
I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!
abstract-algebra group-theory sylow-theory
asked Jan 6 at 22:06
pendermathpendermath
52910
$endgroup$
In a group $G$ of order $400 = 2^4 cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?
I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!
abstract-algebra group-theory sylow-theory
abstract-algebra group-theory sylow-theory
asked Jan 6 at 22:06
pendermathpendermath
52910
asked Jan 6 at 22:06
pendermathpendermath
52910
asked Jan 6 at 22:06
pendermathpendermath
52910
asked Jan 6 at 22:06
pendermathpendermath
52910
asked Jan 6 at 22:06
pendermathpendermath
52910
52910
add a comment |
add a comment |
1 Answer
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$begingroup$
This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
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$begingroup$
This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
add a comment |
$begingroup$
This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
add a comment |
$begingroup$
This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
answered Jan 6 at 22:13
Matt SamuelMatt Samuel
37.7k63665
37.7k63665
add a comment |
add a comment |
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