prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelianNonabelian $p$-groups all of whose proper subgroups are abelian.For what numbers $n$ every group of order $n$ is abelian?Prove that a finite abelian group is simple if and only if its order is prime.A finite group with the property that all of its proper subgroups are abelianProve that $ AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime.Using Lagrange's theorem, prove that a non-abelian group of order $10$ must have a subgroup of order $5$.Which of the following group has a proper subgroup that is not cyclic?Center of Simple Abelian Group and Simple Nonabelian GroupProof verification: Prove whether or not every proper subgroup of a nonabelian group is nonabelian.Show that the order of group $G$ is $1$ or a prime number
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prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian
Nonabelian $p$-groups all of whose proper subgroups are abelian.For what numbers $n$ every group of order $n$ is abelian?Prove that a finite abelian group is simple if and only if its order is prime.A finite group with the property that all of its proper subgroups are abelianProve that $ AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime.Using Lagrange's theorem, prove that a non-abelian group of order $10$ must have a subgroup of order $5$.Which of the following group has a proper subgroup that is not cyclic?Center of Simple Abelian Group and Simple Nonabelian GroupProof verification: Prove whether or not every proper subgroup of a nonabelian group is nonabelian.Show that the order of group $G$ is $1$ or a prime number
$begingroup$
I need to prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian. I use that, from Lagrange's theorem, the order of the proper subgroups must divide the order of the group and so the subgroups have the orders $p,q$. If $(p,q)=1$, then there exist $k,linmathbbZ$ such that $pk+gl=1$ and so, for an $x$ of a subgroup of order $p$ we obtain $x=x^lq$ and similarly, for an $y$ of order $q$ $y=y^pk$. Also, $p|lq-1$ and $q|pk-1$. This are the only valuable information I obtained. How can I prove this?
abstract-algebra group-theory finite-groups abelian-groups cyclic-groups
$endgroup$
add a comment |
$begingroup$
I need to prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian. I use that, from Lagrange's theorem, the order of the proper subgroups must divide the order of the group and so the subgroups have the orders $p,q$. If $(p,q)=1$, then there exist $k,linmathbbZ$ such that $pk+gl=1$ and so, for an $x$ of a subgroup of order $p$ we obtain $x=x^lq$ and similarly, for an $y$ of order $q$ $y=y^pk$. Also, $p|lq-1$ and $q|pk-1$. This are the only valuable information I obtained. How can I prove this?
abstract-algebra group-theory finite-groups abelian-groups cyclic-groups
$endgroup$
$begingroup$
You need just the first line.
$endgroup$
– the_fox
Mar 14 at 8:24
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38
add a comment |
$begingroup$
I need to prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian. I use that, from Lagrange's theorem, the order of the proper subgroups must divide the order of the group and so the subgroups have the orders $p,q$. If $(p,q)=1$, then there exist $k,linmathbbZ$ such that $pk+gl=1$ and so, for an $x$ of a subgroup of order $p$ we obtain $x=x^lq$ and similarly, for an $y$ of order $q$ $y=y^pk$. Also, $p|lq-1$ and $q|pk-1$. This are the only valuable information I obtained. How can I prove this?
abstract-algebra group-theory finite-groups abelian-groups cyclic-groups
$endgroup$
I need to prove that for $G$ a multiplicative nonabelian group of order $pq$, where $p$ and $q$ are prime numbers, any proper subgroup of $G$ is abelian. I use that, from Lagrange's theorem, the order of the proper subgroups must divide the order of the group and so the subgroups have the orders $p,q$. If $(p,q)=1$, then there exist $k,linmathbbZ$ such that $pk+gl=1$ and so, for an $x$ of a subgroup of order $p$ we obtain $x=x^lq$ and similarly, for an $y$ of order $q$ $y=y^pk$. Also, $p|lq-1$ and $q|pk-1$. This are the only valuable information I obtained. How can I prove this?
abstract-algebra group-theory finite-groups abelian-groups cyclic-groups
abstract-algebra group-theory finite-groups abelian-groups cyclic-groups
asked Mar 14 at 7:58
user651754
$begingroup$
You need just the first line.
$endgroup$
– the_fox
Mar 14 at 8:24
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38
add a comment |
$begingroup$
You need just the first line.
$endgroup$
– the_fox
Mar 14 at 8:24
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38
$begingroup$
You need just the first line.
$endgroup$
– the_fox
Mar 14 at 8:24
$begingroup$
You need just the first line.
$endgroup$
– the_fox
Mar 14 at 8:24
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
$endgroup$
add a comment |
$begingroup$
By Lagrange's theorem we can say that order of any proper subgroup of $G$ is either of order $p$ or of order $q.$ Since $p$ and $q$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $G$ is cyclic and hence abelian.
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
$endgroup$
add a comment |
$begingroup$
Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
$endgroup$
add a comment |
$begingroup$
Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
$endgroup$
Hint: use Lagrange's Theorem, the order of a subgroup divides the order of the whole group. And, a group of prime order must be cyclic (also follows from Lagrange).
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
answered Mar 14 at 8:10
Nicky HeksterNicky Hekster
29k63456
29k63456
add a comment |
add a comment |
$begingroup$
By Lagrange's theorem we can say that order of any proper subgroup of $G$ is either of order $p$ or of order $q.$ Since $p$ and $q$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $G$ is cyclic and hence abelian.
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
$endgroup$
add a comment |
$begingroup$
By Lagrange's theorem we can say that order of any proper subgroup of $G$ is either of order $p$ or of order $q.$ Since $p$ and $q$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $G$ is cyclic and hence abelian.
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
$endgroup$
add a comment |
$begingroup$
By Lagrange's theorem we can say that order of any proper subgroup of $G$ is either of order $p$ or of order $q.$ Since $p$ and $q$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $G$ is cyclic and hence abelian.
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
$endgroup$
By Lagrange's theorem we can say that order of any proper subgroup of $G$ is either of order $p$ or of order $q.$ Since $p$ and $q$ are primes and we know that any group of prime order is cyclic so any proper subgroup of the group $G$ is cyclic and hence abelian.
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
answered Mar 14 at 8:24
Dbchatto67Dbchatto67
1,995319
1,995319
add a comment |
add a comment |
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$begingroup$
You need just the first line.
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– the_fox
Mar 14 at 8:24
$begingroup$
Try to avoid unnecessary assumptions. You don't need to assume that the group $G$ is multiplicative, and you don't need to assume that it is nonabelian.
$endgroup$
– Derek Holt
Mar 14 at 8:38