Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable.












0












$begingroup$


I got a problem as below.




Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable. ($p$ and $q$ may or may not be distinct.)




Honestly, I even don't know where to start. Thanks for any help in advance!



Definition of Solvability(From dummit and Foote)



A group $G$ is solvable if there is a chain of subgroups $$1=G_0trianglelefteq G_1trianglelefteq cdots trianglelefteq G_s=G$$such that $G_{i+1}/G_i$ is abelian for $i=0,1,dots, s-1$.










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$endgroup$








  • 1




    $begingroup$
    Is there an index two normal subgroup?
    $endgroup$
    – dan_fulea
    Jan 15 at 23:47










  • $begingroup$
    @dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
    $endgroup$
    – Lev Ban
    Jan 15 at 23:49










  • $begingroup$
    Some elements have odd order, some elements have even order...
    $endgroup$
    – dan_fulea
    Jan 15 at 23:55










  • $begingroup$
    all primes are odd
    $endgroup$
    – clathratus
    Jan 16 at 0:04










  • $begingroup$
    @clathratus the prime 2?
    $endgroup$
    – Nico
    Jan 16 at 0:09
















0












$begingroup$


I got a problem as below.




Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable. ($p$ and $q$ may or may not be distinct.)




Honestly, I even don't know where to start. Thanks for any help in advance!



Definition of Solvability(From dummit and Foote)



A group $G$ is solvable if there is a chain of subgroups $$1=G_0trianglelefteq G_1trianglelefteq cdots trianglelefteq G_s=G$$such that $G_{i+1}/G_i$ is abelian for $i=0,1,dots, s-1$.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is there an index two normal subgroup?
    $endgroup$
    – dan_fulea
    Jan 15 at 23:47










  • $begingroup$
    @dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
    $endgroup$
    – Lev Ban
    Jan 15 at 23:49










  • $begingroup$
    Some elements have odd order, some elements have even order...
    $endgroup$
    – dan_fulea
    Jan 15 at 23:55










  • $begingroup$
    all primes are odd
    $endgroup$
    – clathratus
    Jan 16 at 0:04










  • $begingroup$
    @clathratus the prime 2?
    $endgroup$
    – Nico
    Jan 16 at 0:09














0












0








0


1



$begingroup$


I got a problem as below.




Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable. ($p$ and $q$ may or may not be distinct.)




Honestly, I even don't know where to start. Thanks for any help in advance!



Definition of Solvability(From dummit and Foote)



A group $G$ is solvable if there is a chain of subgroups $$1=G_0trianglelefteq G_1trianglelefteq cdots trianglelefteq G_s=G$$such that $G_{i+1}/G_i$ is abelian for $i=0,1,dots, s-1$.










share|cite|improve this question











$endgroup$




I got a problem as below.




Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable. ($p$ and $q$ may or may not be distinct.)




Honestly, I even don't know where to start. Thanks for any help in advance!



Definition of Solvability(From dummit and Foote)



A group $G$ is solvable if there is a chain of subgroups $$1=G_0trianglelefteq G_1trianglelefteq cdots trianglelefteq G_s=G$$such that $G_{i+1}/G_i$ is abelian for $i=0,1,dots, s-1$.







abstract-algebra group-theory prime-numbers finite-groups






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 0:20







Lev Ban

















asked Jan 15 at 23:43









Lev BanLev Ban

1,0751317




1,0751317








  • 1




    $begingroup$
    Is there an index two normal subgroup?
    $endgroup$
    – dan_fulea
    Jan 15 at 23:47










  • $begingroup$
    @dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
    $endgroup$
    – Lev Ban
    Jan 15 at 23:49










  • $begingroup$
    Some elements have odd order, some elements have even order...
    $endgroup$
    – dan_fulea
    Jan 15 at 23:55










  • $begingroup$
    all primes are odd
    $endgroup$
    – clathratus
    Jan 16 at 0:04










  • $begingroup$
    @clathratus the prime 2?
    $endgroup$
    – Nico
    Jan 16 at 0:09














  • 1




    $begingroup$
    Is there an index two normal subgroup?
    $endgroup$
    – dan_fulea
    Jan 15 at 23:47










  • $begingroup$
    @dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
    $endgroup$
    – Lev Ban
    Jan 15 at 23:49










  • $begingroup$
    Some elements have odd order, some elements have even order...
    $endgroup$
    – dan_fulea
    Jan 15 at 23:55










  • $begingroup$
    all primes are odd
    $endgroup$
    – clathratus
    Jan 16 at 0:04










  • $begingroup$
    @clathratus the prime 2?
    $endgroup$
    – Nico
    Jan 16 at 0:09








1




1




$begingroup$
Is there an index two normal subgroup?
$endgroup$
– dan_fulea
Jan 15 at 23:47




$begingroup$
Is there an index two normal subgroup?
$endgroup$
– dan_fulea
Jan 15 at 23:47












$begingroup$
@dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
$endgroup$
– Lev Ban
Jan 15 at 23:49




$begingroup$
@dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
$endgroup$
– Lev Ban
Jan 15 at 23:49












$begingroup$
Some elements have odd order, some elements have even order...
$endgroup$
– dan_fulea
Jan 15 at 23:55




$begingroup$
Some elements have odd order, some elements have even order...
$endgroup$
– dan_fulea
Jan 15 at 23:55












$begingroup$
all primes are odd
$endgroup$
– clathratus
Jan 16 at 0:04




$begingroup$
all primes are odd
$endgroup$
– clathratus
Jan 16 at 0:04












$begingroup$
@clathratus the prime 2?
$endgroup$
– Nico
Jan 16 at 0:09




$begingroup$
@clathratus the prime 2?
$endgroup$
– Nico
Jan 16 at 0:09










1 Answer
1






active

oldest

votes


















4












$begingroup$

Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.



Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$)



If $p=qne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_pequiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_rne 1$, which implies $n_rin{p,p^2}$. But if $n_r=p$ then we'll get $pequiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)



Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_qne r$ and hence $n_qgeq p$. And finally $n_rgeq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:



$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)geq pqr+1$



Which is of course a contradiction.



Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $Ntriangleleft G$ which is not $G$ and not ${e}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:34








  • 1




    $begingroup$
    They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
    $endgroup$
    – Mark
    Jan 16 at 0:37










  • $begingroup$
    You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
    $endgroup$
    – Mark
    Jan 16 at 0:39










  • $begingroup$
    Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:41






  • 1




    $begingroup$
    A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
    $endgroup$
    – Mark
    Jan 16 at 0:46














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1 Answer
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1 Answer
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active

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active

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active

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4












$begingroup$

Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.



Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$)



If $p=qne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_pequiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_rne 1$, which implies $n_rin{p,p^2}$. But if $n_r=p$ then we'll get $pequiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)



Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_qne r$ and hence $n_qgeq p$. And finally $n_rgeq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:



$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)geq pqr+1$



Which is of course a contradiction.



Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $Ntriangleleft G$ which is not $G$ and not ${e}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:34








  • 1




    $begingroup$
    They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
    $endgroup$
    – Mark
    Jan 16 at 0:37










  • $begingroup$
    You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
    $endgroup$
    – Mark
    Jan 16 at 0:39










  • $begingroup$
    Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:41






  • 1




    $begingroup$
    A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
    $endgroup$
    – Mark
    Jan 16 at 0:46


















4












$begingroup$

Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.



Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$)



If $p=qne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_pequiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_rne 1$, which implies $n_rin{p,p^2}$. But if $n_r=p$ then we'll get $pequiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)



Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_qne r$ and hence $n_qgeq p$. And finally $n_rgeq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:



$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)geq pqr+1$



Which is of course a contradiction.



Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $Ntriangleleft G$ which is not $G$ and not ${e}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:34








  • 1




    $begingroup$
    They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
    $endgroup$
    – Mark
    Jan 16 at 0:37










  • $begingroup$
    You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
    $endgroup$
    – Mark
    Jan 16 at 0:39










  • $begingroup$
    Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:41






  • 1




    $begingroup$
    A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
    $endgroup$
    – Mark
    Jan 16 at 0:46
















4












4








4





$begingroup$

Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.



Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$)



If $p=qne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_pequiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_rne 1$, which implies $n_rin{p,p^2}$. But if $n_r=p$ then we'll get $pequiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)



Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_qne r$ and hence $n_qgeq p$. And finally $n_rgeq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:



$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)geq pqr+1$



Which is of course a contradiction.



Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $Ntriangleleft G$ which is not $G$ and not ${e}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable.






share|cite|improve this answer











$endgroup$



Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.



Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$)



If $p=qne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_pequiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_rne 1$, which implies $n_rin{p,p^2}$. But if $n_r=p$ then we'll get $pequiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)



Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_qne r$ and hence $n_qgeq p$. And finally $n_rgeq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:



$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)geq pqr+1$



Which is of course a contradiction.



Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $Ntriangleleft G$ which is not $G$ and not ${e}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered Jan 16 at 0:14









MarkMark

11.5k1824




11.5k1824












  • $begingroup$
    In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:34








  • 1




    $begingroup$
    They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
    $endgroup$
    – Mark
    Jan 16 at 0:37










  • $begingroup$
    You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
    $endgroup$
    – Mark
    Jan 16 at 0:39










  • $begingroup$
    Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:41






  • 1




    $begingroup$
    A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
    $endgroup$
    – Mark
    Jan 16 at 0:46




















  • $begingroup$
    In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:34








  • 1




    $begingroup$
    They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
    $endgroup$
    – Mark
    Jan 16 at 0:37










  • $begingroup$
    You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
    $endgroup$
    – Mark
    Jan 16 at 0:39










  • $begingroup$
    Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
    $endgroup$
    – Lev Ban
    Jan 16 at 0:41






  • 1




    $begingroup$
    A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
    $endgroup$
    – Mark
    Jan 16 at 0:46


















$begingroup$
In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
$endgroup$
– Lev Ban
Jan 16 at 0:34






$begingroup$
In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
$endgroup$
– Lev Ban
Jan 16 at 0:34






1




1




$begingroup$
They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
$endgroup$
– Mark
Jan 16 at 0:37




$begingroup$
They are not necessary abelian. I'm just using a very basic theorem: if $G$ has a normal subgroup $N$ such that $N$ and $G/N$ are both solvable then $G$ is solvable. This is probably the most important thing you should know about solvable groups.
$endgroup$
– Mark
Jan 16 at 0:37












$begingroup$
You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
$endgroup$
– Mark
Jan 16 at 0:39




$begingroup$
You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
$endgroup$
– Mark
Jan 16 at 0:39












$begingroup$
Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
$endgroup$
– Lev Ban
Jan 16 at 0:41




$begingroup$
Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
$endgroup$
– Lev Ban
Jan 16 at 0:41




1




1




$begingroup$
A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
$endgroup$
– Mark
Jan 16 at 0:46






$begingroup$
A group of prime order is abelian, hence solvable. As for a product of two primes you can easily prove that a group of order $pq$ is not simple. (using Sylow theorems: it is a much easier proof compared to the proof of $pqr$ which I did). Hence if $G$ is a group of order $pq$ then it has a normal subgroup $N$ with order $p$ or $q$. Both $N$ and $G/N$ are of prime order in this case, hence solvable. And again, by the theorem I already used such $G$ is solvable as well.
$endgroup$
– Mark
Jan 16 at 0:46




















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