Groups of order $360$ have a subgroup of order $10$
$begingroup$
I want to prove that groups of order $360$ must have a subgroup of order $10$.
By Sylow's theorem, the number of Sylow $5$-subgroups $n_5 equiv 1 pmod 5$ and $n_5mid 360$. There are three solutions: $1, 6, 36$ (let me know if I missed any).
If $n_5=1$, then the only one is normal, making the product with an element of order $2$ we get a subgroup of order $10$.
If $n_5=36$, then pick any Sylow $5$-subgroup, $[G:N_{G}(P)]=36$. It follows that $N_G(P)$ is a subgroup of order $10$.
But how to deal with the case when $n_5=6$?
abstract-algebra group-theory finite-groups
edited Jan 11 at 12:37
the_fox
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
$endgroup$
add a comment |
$begingroup$
I want to prove that groups of order $360$ must have a subgroup of order $10$.
By Sylow's theorem, the number of Sylow $5$-subgroups $n_5 equiv 1 pmod 5$ and $n_5mid 360$. There are three solutions: $1, 6, 36$ (let me know if I missed any).
If $n_5=1$, then the only one is normal, making the product with an element of order $2$ we get a subgroup of order $10$.
If $n_5=36$, then pick any Sylow $5$-subgroup, $[G:N_{G}(P)]=36$. It follows that $N_G(P)$ is a subgroup of order $10$.
But how to deal with the case when $n_5=6$?
abstract-algebra group-theory finite-groups
edited Jan 11 at 12:37
the_fox
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
$endgroup$
add a comment |
$begingroup$
I want to prove that groups of order $360$ must have a subgroup of order $10$.
By Sylow's theorem, the number of Sylow $5$-subgroups $n_5 equiv 1 pmod 5$ and $n_5mid 360$. There are three solutions: $1, 6, 36$ (let me know if I missed any).
If $n_5=1$, then the only one is normal, making the product with an element of order $2$ we get a subgroup of order $10$.
If $n_5=36$, then pick any Sylow $5$-subgroup, $[G:N_{G}(P)]=36$. It follows that $N_G(P)$ is a subgroup of order $10$.
But how to deal with the case when $n_5=6$?
abstract-algebra group-theory finite-groups
edited Jan 11 at 12:37
the_fox
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
$endgroup$
I want to prove that groups of order $360$ must have a subgroup of order $10$.
By Sylow's theorem, the number of Sylow $5$-subgroups $n_5 equiv 1 pmod 5$ and $n_5mid 360$. There are three solutions: $1, 6, 36$ (let me know if I missed any).
If $n_5=1$, then the only one is normal, making the product with an element of order $2$ we get a subgroup of order $10$.
If $n_5=36$, then pick any Sylow $5$-subgroup, $[G:N_{G}(P)]=36$. It follows that $N_G(P)$ is a subgroup of order $10$.
But how to deal with the case when $n_5=6$?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 11 at 12:37
the_fox
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
edited Jan 11 at 12:37
the_fox
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
edited Jan 11 at 12:37
the_fox
2,58711533
edited Jan 11 at 12:37
the_fox
2,58711533
edited Jan 11 at 12:37
the_fox
2,58711533
2,58711533
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
asked Sep 9 '18 at 19:17
No OneNo One
2,0241519
2,0241519
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work.
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)leq Hleq G$ is a subgroup of order $10$.
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
$endgroup$
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work.
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)leq Hleq G$ is a subgroup of order $10$.
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
$endgroup$
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
add a comment |
$begingroup$
If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work.
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)leq Hleq G$ is a subgroup of order $10$.
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
$endgroup$
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
add a comment |
$begingroup$
If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work.
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)leq Hleq G$ is a subgroup of order $10$.
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
$endgroup$
If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work.
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)leq Hleq G$ is a subgroup of order $10$.
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
answered Sep 9 '18 at 19:35
SMMSMM
2,21859
2,21859
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
add a comment |
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
$begingroup$
You already know that $n_5=1$ in $H$ since it has a normal Sylow subgroup.
$endgroup$
– Levent
Jan 11 at 13:08
add a comment |
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