proving that a group of order 60 is simple using homomorphisms.
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I want to show that the group G, where $|G|=60$ and has 20 elements of order 3 is simple.
Here's what I did:
Suppose that G is not simple this implies that $n_3>1, n_2>1, n_5>1$ where these n's denote the number of sylow p-subgroups in G.
Consider $n_5=(1+5k)|12 Rightarrow n_5=6$ (as we've said it cant be one)
But then $|G:N_G(P)|=6$ ( P denoting the sylow 5 subgroups here).
This implies there is a homomorphism induced by G acting on the left cosets of G/P by left multiplication. $|G/P|>1$ and operation is transitive so the kernel is not the whole group. since g is not simple the map is not injective and also since the groups are finite not surjective. which implies $|G|>|S_n|$ a contradiction .....so G is simple.
Is this correct ?
If not what specifically is wrong with it ?
group-theory proof-verification proof-writing finite-groups sylow-theory
asked Nov 14 at 1:51
exodius
924417
add a comment |
up vote
0
down vote
favorite
I want to show that the group G, where $|G|=60$ and has 20 elements of order 3 is simple.
Here's what I did:
Suppose that G is not simple this implies that $n_3>1, n_2>1, n_5>1$ where these n's denote the number of sylow p-subgroups in G.
Consider $n_5=(1+5k)|12 Rightarrow n_5=6$ (as we've said it cant be one)
But then $|G:N_G(P)|=6$ ( P denoting the sylow 5 subgroups here).
This implies there is a homomorphism induced by G acting on the left cosets of G/P by left multiplication. $|G/P|>1$ and operation is transitive so the kernel is not the whole group. since g is not simple the map is not injective and also since the groups are finite not surjective. which implies $|G|>|S_n|$ a contradiction .....so G is simple.
Is this correct ?
If not what specifically is wrong with it ?
group-theory proof-verification proof-writing finite-groups sylow-theory
asked Nov 14 at 1:51
exodius
924417
"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
1
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show that the group G, where $|G|=60$ and has 20 elements of order 3 is simple.
Here's what I did:
Suppose that G is not simple this implies that $n_3>1, n_2>1, n_5>1$ where these n's denote the number of sylow p-subgroups in G.
Consider $n_5=(1+5k)|12 Rightarrow n_5=6$ (as we've said it cant be one)
But then $|G:N_G(P)|=6$ ( P denoting the sylow 5 subgroups here).
This implies there is a homomorphism induced by G acting on the left cosets of G/P by left multiplication. $|G/P|>1$ and operation is transitive so the kernel is not the whole group. since g is not simple the map is not injective and also since the groups are finite not surjective. which implies $|G|>|S_n|$ a contradiction .....so G is simple.
Is this correct ?
If not what specifically is wrong with it ?
group-theory proof-verification proof-writing finite-groups sylow-theory
asked Nov 14 at 1:51
exodius
924417
I want to show that the group G, where $|G|=60$ and has 20 elements of order 3 is simple.
Here's what I did:
Suppose that G is not simple this implies that $n_3>1, n_2>1, n_5>1$ where these n's denote the number of sylow p-subgroups in G.
Consider $n_5=(1+5k)|12 Rightarrow n_5=6$ (as we've said it cant be one)
But then $|G:N_G(P)|=6$ ( P denoting the sylow 5 subgroups here).
This implies there is a homomorphism induced by G acting on the left cosets of G/P by left multiplication. $|G/P|>1$ and operation is transitive so the kernel is not the whole group. since g is not simple the map is not injective and also since the groups are finite not surjective. which implies $|G|>|S_n|$ a contradiction .....so G is simple.
Is this correct ?
If not what specifically is wrong with it ?
group-theory proof-verification proof-writing finite-groups sylow-theory
group-theory proof-verification proof-writing finite-groups sylow-theory
asked Nov 14 at 1:51
exodius
924417
asked Nov 14 at 1:51
exodius
924417
edited Nov 14 at 2:27
asked Nov 14 at 1:51
exodius
924417
asked Nov 14 at 1:51
exodius
924417
asked Nov 14 at 1:51
exodius
924417
924417
"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
1
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59
add a comment |
"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
1
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59
"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
1
1
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59
add a comment |
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"since $G$ is not simple the map is not injective..." - How does that follow?
– Bungo
Nov 14 at 2:51
Actually you are assuming not simple, that means there is some normal subgroup either of order 4 or 5
– Shubham
Nov 14 at 2:55
@Bungo I know that If G were simple then it would be an injective map. maybe what is the true statement is the map may not be injective ?
– exodius
Nov 14 at 2:58
1
Simple $implies$ injective, but that doesn't mean that (not simple) $implies$ (not injective), or equivalently, that injective $implies$ simple.
– Bungo
Nov 14 at 2:59