Let $G=langle (1,2,3,4,5,6,7), (2,4,3,7,5,6) rangle$

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Let $G=langle (1,2,3,4,5,6,7), (2,4,3,7,5,6) rangle$.



I need to find sylow basis of $G$.



since the order of $(1,2,3,4,5,6,7)$ is 7 so we can prove that $n_7=1$ then it means that $langle (1,2,3,4,5,6,7)ranglelhd G$ right? this is in the basis but how to I find the rest of the groups in the sylow basis?



I tried to find more groups in the basis, I thoght about the group $langle (2,4,3,7,5,6)rangle$ but she is from order 6 and we need from prime order. But I don't know how to find subgroups from order 2 and 3.



please help if you know how




finite-groups sylow-theory




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  • 1




    $b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
    – daruma
    Aug 3 at 13:11










  • I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
    – Rimon
    Aug 3 at 14:44










  • @daruma , am I right? this is the sylow basis?
    – Rimon
    Aug 3 at 20:00














up vote
0
down vote

favorite












Let $G=langle (1,2,3,4,5,6,7), (2,4,3,7,5,6) rangle$.



I need to find sylow basis of $G$.



since the order of $(1,2,3,4,5,6,7)$ is 7 so we can prove that $n_7=1$ then it means that $langle (1,2,3,4,5,6,7)ranglelhd G$ right? this is in the basis but how to I find the rest of the groups in the sylow basis?



I tried to find more groups in the basis, I thoght about the group $langle (2,4,3,7,5,6)rangle$ but she is from order 6 and we need from prime order. But I don't know how to find subgroups from order 2 and 3.



please help if you know how




finite-groups sylow-theory




share|cite|improve this question















  • 1




    $b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
    – daruma
    Aug 3 at 13:11










  • I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
    – Rimon
    Aug 3 at 14:44










  • @daruma , am I right? this is the sylow basis?
    – Rimon
    Aug 3 at 20:00












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $G=langle (1,2,3,4,5,6,7), (2,4,3,7,5,6) rangle$.



I need to find sylow basis of $G$.



since the order of $(1,2,3,4,5,6,7)$ is 7 so we can prove that $n_7=1$ then it means that $langle (1,2,3,4,5,6,7)ranglelhd G$ right? this is in the basis but how to I find the rest of the groups in the sylow basis?



I tried to find more groups in the basis, I thoght about the group $langle (2,4,3,7,5,6)rangle$ but she is from order 6 and we need from prime order. But I don't know how to find subgroups from order 2 and 3.



please help if you know how




finite-groups sylow-theory




share|cite|improve this question











Let $G=langle (1,2,3,4,5,6,7), (2,4,3,7,5,6) rangle$.



I need to find sylow basis of $G$.



since the order of $(1,2,3,4,5,6,7)$ is 7 so we can prove that $n_7=1$ then it means that $langle (1,2,3,4,5,6,7)ranglelhd G$ right? this is in the basis but how to I find the rest of the groups in the sylow basis?



I tried to find more groups in the basis, I thoght about the group $langle (2,4,3,7,5,6)rangle$ but she is from order 6 and we need from prime order. But I don't know how to find subgroups from order 2 and 3.



please help if you know how





finite-groups sylow-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 3 at 13:02









Rimon

33




33







  • 1




    $b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
    – daruma
    Aug 3 at 13:11










  • I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
    – Rimon
    Aug 3 at 14:44










  • @daruma , am I right? this is the sylow basis?
    – Rimon
    Aug 3 at 20:00












  • 1




    $b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
    – daruma
    Aug 3 at 13:11










  • I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
    – Rimon
    Aug 3 at 14:44










  • @daruma , am I right? this is the sylow basis?
    – Rimon
    Aug 3 at 20:00







1




1




$b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
– daruma
Aug 3 at 13:11




$b=(2,4,3,7,5,6)$ is an element of order $6$ so $b^3$ generates a subgroup of order $2$ and $b^2$ generates a subgroup of order $3$. But how do you know there aren't Sylow-2 groups and Sylow-3 groups of higher power? You need to prove $|G|=7times 6$.
– daruma
Aug 3 at 13:11












I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
– Rimon
Aug 3 at 14:44




I know how to prove that the order is 42 :) then the sylow basis is $langle(1,2,3,4,5,6,7)rangle, langle b^3 rangle, langle b^2 rangle$?
– Rimon
Aug 3 at 14:44












@daruma , am I right? this is the sylow basis?
– Rimon
Aug 3 at 20:00




@daruma , am I right? this is the sylow basis?
– Rimon
Aug 3 at 20:00















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