$G$ is a group of order $60$. Will $G$ always contain a subgroup of order $6$?
$begingroup$
$G$ is a group of order $60$. Will there be a subgroup of order $ 6$?
Alternating group $A_5$ has a subgroup of order $6$. That is the group generated by this set ${(123), (23) (45)}$.
Will we be able to prove that there always exists a subgroup of order $6$ in a group of order $ 60$?
Can anyone help me to understand by giving a hint?
abstract-algebra group-theory finite-groups
edited Jan 11 at 7:22
the_fox
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
$endgroup$
add a comment |
$begingroup$
$G$ is a group of order $60$. Will there be a subgroup of order $ 6$?
Alternating group $A_5$ has a subgroup of order $6$. That is the group generated by this set ${(123), (23) (45)}$.
Will we be able to prove that there always exists a subgroup of order $6$ in a group of order $ 60$?
Can anyone help me to understand by giving a hint?
abstract-algebra group-theory finite-groups
edited Jan 11 at 7:22
the_fox
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
$endgroup$
$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
1
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23
add a comment |
$begingroup$
$G$ is a group of order $60$. Will there be a subgroup of order $ 6$?
Alternating group $A_5$ has a subgroup of order $6$. That is the group generated by this set ${(123), (23) (45)}$.
Will we be able to prove that there always exists a subgroup of order $6$ in a group of order $ 60$?
Can anyone help me to understand by giving a hint?
abstract-algebra group-theory finite-groups
edited Jan 11 at 7:22
the_fox
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
$endgroup$
$G$ is a group of order $60$. Will there be a subgroup of order $ 6$?
Alternating group $A_5$ has a subgroup of order $6$. That is the group generated by this set ${(123), (23) (45)}$.
Will we be able to prove that there always exists a subgroup of order $6$ in a group of order $ 60$?
Can anyone help me to understand by giving a hint?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 11 at 7:22
the_fox
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
edited Jan 11 at 7:22
the_fox
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
edited Jan 11 at 7:22
the_fox
2,58711533
edited Jan 11 at 7:22
the_fox
2,58711533
edited Jan 11 at 7:22
the_fox
2,58711533
2,58711533
asked Jan 11 at 5:12
cmicmi
1,109212
asked Jan 11 at 5:12
cmicmi
1,109212
asked Jan 11 at 5:12
cmicmi
1,109212
1,109212
$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
1
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23
add a comment |
$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
1
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23
$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
1
1
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, that's not always true. Take for example $G = C_5 times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
$endgroup$
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
add a comment |
Your Answer
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1 Answer
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$begingroup$
No, that's not always true. Take for example $G = C_5 times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
$endgroup$
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
add a comment |
$begingroup$
No, that's not always true. Take for example $G = C_5 times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
$endgroup$
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
add a comment |
$begingroup$
No, that's not always true. Take for example $G = C_5 times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
$endgroup$
No, that's not always true. Take for example $G = C_5 times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
answered Jan 11 at 5:24
the_foxthe_fox
2,58711533
2,58711533
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
add a comment |
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
$begingroup$
$C_5$ means?@the_fox
$endgroup$
– cmi
Jan 11 at 5:45
1
1
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
Cyclic group of order $5$.
$endgroup$
– the_fox
Jan 11 at 5:46
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
$begingroup$
No I can not see why it is the only exception @the_fox
$endgroup$
– cmi
Jan 11 at 5:57
1
1
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
$begingroup$
Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$?
$endgroup$
– the_fox
Jan 11 at 6:07
add a comment |
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$begingroup$
Maybe some similar questions will inspire you: math.stackexchange.com/q/94548 math.stackexchange.com/q/2594060 math.stackexchange.com/q/2911121
$endgroup$
– Trevor Gunn
Jan 11 at 5:18
$begingroup$
Okk I am trying to get some idea from them@TrevorGunn
$endgroup$
– cmi
Jan 11 at 5:20
1
$begingroup$
Use Sylow theorems. en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups.
$endgroup$
– Pratyush Sarkar
Jan 11 at 5:23