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 ?










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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

















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 ?










share|cite|improve this question
























  • "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















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 ?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 14 at 2:27

























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




















  • "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

















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