Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable.
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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
$endgroup$
|
show 1 more comment
$begingroup$
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 $i=0,1,dots, s-1$.
abstract-algebra group-theory prime-numbers finite-groups
$endgroup$
1
$begingroup$
Is there an index two normal subgroup?
$endgroup$
– dan_fulea
Jan 15 at 23:47
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@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
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@clathratus the prime 2?
$endgroup$
– Nico
Jan 16 at 0:09
|
show 1 more comment
$begingroup$
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 $i=0,1,dots, s-1$.
abstract-algebra group-theory prime-numbers finite-groups
$endgroup$
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
abstract-algebra group-theory prime-numbers finite-groups
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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.
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– Mark
Jan 16 at 0:46
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In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
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– Lev Ban
Jan 16 at 0:34
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In you case, the chain is $$1trianglelefteq N trianglelefteq G.$$ Am I right? Then why $G/N$ and $N$ are abelian?
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– Lev Ban
Jan 16 at 0:34
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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.
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– Mark
Jan 16 at 0:37
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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.
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– Mark
Jan 16 at 0:37
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You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
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– Mark
Jan 16 at 0:39
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You can see a proof of this theorem here: math.stackexchange.com/questions/933146/…
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– Mark
Jan 16 at 0:39
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Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
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– Lev Ban
Jan 16 at 0:41
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Thanks! Could you give me more explanation why such $N$ and $G/N$ with such order are solvable?
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– 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.
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– Mark
Jan 16 at 0:46
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1
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Is there an index two normal subgroup?
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– dan_fulea
Jan 15 at 23:47
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@dan_fulea All the given condition is that $p$ and $q$ are odd prime. No more than that :(
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– Lev Ban
Jan 15 at 23:49
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Some elements have odd order, some elements have even order...
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– dan_fulea
Jan 15 at 23:55
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all primes are odd
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– clathratus
Jan 16 at 0:04
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@clathratus the prime 2?
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– Nico
Jan 16 at 0:09