Proving that there exists a subgroup of $G/N$ that's isomorphic to $Hle G, text{with} |H|=p$ and $Hnleq N$
Let $G$ be a group with a normal subgroup $N$ and a subgroup $H$ with $|H|=p$ of prime cardinality and $H$ is not a subgroup of $N$.
The question is to prove that $G/N$ has a subgroup that is isomorphic to $H$.
My attempt:
$|H|=pimplies $ $H$ is cyclic. $H=langle hrangle$ for some $hin H$
The correspondance theorem states that $f:{Kle G : Ksupset N}rightarrow{text{subgroups of } G/N}$ is a bijection but this doesn't seem to be useful here.
A hint would be welcome
abstract-algebra group-theory
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
add a comment |
Let $G$ be a group with a normal subgroup $N$ and a subgroup $H$ with $|H|=p$ of prime cardinality and $H$ is not a subgroup of $N$.
The question is to prove that $G/N$ has a subgroup that is isomorphic to $H$.
My attempt:
$|H|=pimplies $ $H$ is cyclic. $H=langle hrangle$ for some $hin H$
The correspondance theorem states that $f:{Kle G : Ksupset N}rightarrow{text{subgroups of } G/N}$ is a bijection but this doesn't seem to be useful here.
A hint would be welcome
abstract-algebra group-theory
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
1
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53
add a comment |
Let $G$ be a group with a normal subgroup $N$ and a subgroup $H$ with $|H|=p$ of prime cardinality and $H$ is not a subgroup of $N$.
The question is to prove that $G/N$ has a subgroup that is isomorphic to $H$.
My attempt:
$|H|=pimplies $ $H$ is cyclic. $H=langle hrangle$ for some $hin H$
The correspondance theorem states that $f:{Kle G : Ksupset N}rightarrow{text{subgroups of } G/N}$ is a bijection but this doesn't seem to be useful here.
A hint would be welcome
abstract-algebra group-theory
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
Let $G$ be a group with a normal subgroup $N$ and a subgroup $H$ with $|H|=p$ of prime cardinality and $H$ is not a subgroup of $N$.
The question is to prove that $G/N$ has a subgroup that is isomorphic to $H$.
My attempt:
$|H|=pimplies $ $H$ is cyclic. $H=langle hrangle$ for some $hin H$
The correspondance theorem states that $f:{Kle G : Ksupset N}rightarrow{text{subgroups of } G/N}$ is a bijection but this doesn't seem to be useful here.
A hint would be welcome
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
asked Nov 28 '18 at 11:51
John Cataldo
1,0061216
1,0061216
1
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53
add a comment |
1
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53
1
1
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53
add a comment |
4 Answers
4
active
oldest
votes
Hint:
$N=${$x^p|forall xin G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define $phi: Hto G/N$
H is cyclic group
say H$=<a>$
$phi(a^i)=a^iN$
Show map is well defined, Bijective
answered Nov 28 '18 at 12:10
Shubham
1,5951519
add a comment |
Since $H$ is not contained in $N$, there exists a nonidentity element $gin Hsetminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
answered Nov 28 '18 at 12:12
Hebe
30818
add a comment |
Hint:Let $H=langle hrangle$ for some $hin H$. Define $Psubseteq G/N$ such that $P={h^{m}N mid min Bbb{Z}}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $psi: P to H$ such that $psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
add a comment |
The point is that since $H$ is not a subgroup of $N$, in fact $H cap N = {e}$, since if $h^r in N$ for some $r$, then unless $r equiv 0 mod p$ it has a multiplicative inverse modulo $p$ say $rr' equiv 1 mod p$, in which case $(h^r)^{r'} = h in N$, so $H$ will be a subgroup of $N$,a contradiction.
Now, we may proceed in this way : consider the cyclic group $langle hNrangle subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r in N$, which is a contradiction as we have $h^r in H$, and showed earlier that $H cap N = {e}$. So, the cyclic subgroup $langle hNrangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N cong H /(H cap N)$, but we have already seen that $H cap N = {e}$, so $HN/N cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
$N=${$x^p|forall xin G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define $phi: Hto G/N$
H is cyclic group
say H$=<a>$
$phi(a^i)=a^iN$
Show map is well defined, Bijective
answered Nov 28 '18 at 12:10
Shubham
1,5951519
add a comment |
Hint:
$N=${$x^p|forall xin G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define $phi: Hto G/N$
H is cyclic group
say H$=<a>$
$phi(a^i)=a^iN$
Show map is well defined, Bijective
answered Nov 28 '18 at 12:10
Shubham
1,5951519
add a comment |
Hint:
$N=${$x^p|forall xin G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define $phi: Hto G/N$
H is cyclic group
say H$=<a>$
$phi(a^i)=a^iN$
Show map is well defined, Bijective
answered Nov 28 '18 at 12:10
Shubham
1,5951519
Hint:
$N=${$x^p|forall xin G$}
Step 1: Show N is normal subgroup of G and does not contain element of order p .
Step 2:
Define $phi: Hto G/N$
H is cyclic group
say H$=<a>$
$phi(a^i)=a^iN$
Show map is well defined, Bijective
answered Nov 28 '18 at 12:10
Shubham
1,5951519
answered Nov 28 '18 at 12:10
Shubham
1,5951519
answered Nov 28 '18 at 12:10
Shubham
1,5951519
answered Nov 28 '18 at 12:10
Shubham
1,5951519
1,5951519
add a comment |
add a comment |
Since $H$ is not contained in $N$, there exists a nonidentity element $gin Hsetminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
answered Nov 28 '18 at 12:12
Hebe
30818
add a comment |
Since $H$ is not contained in $N$, there exists a nonidentity element $gin Hsetminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
answered Nov 28 '18 at 12:12
Hebe
30818
add a comment |
Since $H$ is not contained in $N$, there exists a nonidentity element $gin Hsetminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
answered Nov 28 '18 at 12:12
Hebe
30818
Since $H$ is not contained in $N$, there exists a nonidentity element $gin Hsetminus N$. It follows that $gN$ is not identity in $G/N$. Because the order of $g$ is $p$, $(gN)^p=1$ in $G/N$. Hence, the order of $gN$ in $G/N$ divides $p$, and the order must be $p$. Now, the subgroup of $G/N$ generated by $gN$ has order $p$, which is surely isomorphic to $H$.
answered Nov 28 '18 at 12:12
Hebe
30818
answered Nov 28 '18 at 12:12
Hebe
30818
answered Nov 28 '18 at 12:12
Hebe
30818
answered Nov 28 '18 at 12:12
Hebe
30818
30818
add a comment |
add a comment |
Hint:Let $H=langle hrangle$ for some $hin H$. Define $Psubseteq G/N$ such that $P={h^{m}N mid min Bbb{Z}}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $psi: P to H$ such that $psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
add a comment |
Hint:Let $H=langle hrangle$ for some $hin H$. Define $Psubseteq G/N$ such that $P={h^{m}N mid min Bbb{Z}}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $psi: P to H$ such that $psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
add a comment |
Hint:Let $H=langle hrangle$ for some $hin H$. Define $Psubseteq G/N$ such that $P={h^{m}N mid min Bbb{Z}}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $psi: P to H$ such that $psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
Hint:Let $H=langle hrangle$ for some $hin H$. Define $Psubseteq G/N$ such that $P={h^{m}N mid min Bbb{Z}}$.First prove that $P$ is a subgroup of $G/N$. After that, define a map $psi: P to H$ such that $psi(h^mN)=h^m$. Then show the map is a well-defined homomorphism and a bijection.
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,516216
1,516216
add a comment |
add a comment |
The point is that since $H$ is not a subgroup of $N$, in fact $H cap N = {e}$, since if $h^r in N$ for some $r$, then unless $r equiv 0 mod p$ it has a multiplicative inverse modulo $p$ say $rr' equiv 1 mod p$, in which case $(h^r)^{r'} = h in N$, so $H$ will be a subgroup of $N$,a contradiction.
Now, we may proceed in this way : consider the cyclic group $langle hNrangle subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r in N$, which is a contradiction as we have $h^r in H$, and showed earlier that $H cap N = {e}$. So, the cyclic subgroup $langle hNrangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N cong H /(H cap N)$, but we have already seen that $H cap N = {e}$, so $HN/N cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
The point is that since $H$ is not a subgroup of $N$, in fact $H cap N = {e}$, since if $h^r in N$ for some $r$, then unless $r equiv 0 mod p$ it has a multiplicative inverse modulo $p$ say $rr' equiv 1 mod p$, in which case $(h^r)^{r'} = h in N$, so $H$ will be a subgroup of $N$,a contradiction.
Now, we may proceed in this way : consider the cyclic group $langle hNrangle subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r in N$, which is a contradiction as we have $h^r in H$, and showed earlier that $H cap N = {e}$. So, the cyclic subgroup $langle hNrangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N cong H /(H cap N)$, but we have already seen that $H cap N = {e}$, so $HN/N cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
The point is that since $H$ is not a subgroup of $N$, in fact $H cap N = {e}$, since if $h^r in N$ for some $r$, then unless $r equiv 0 mod p$ it has a multiplicative inverse modulo $p$ say $rr' equiv 1 mod p$, in which case $(h^r)^{r'} = h in N$, so $H$ will be a subgroup of $N$,a contradiction.
Now, we may proceed in this way : consider the cyclic group $langle hNrangle subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r in N$, which is a contradiction as we have $h^r in H$, and showed earlier that $H cap N = {e}$. So, the cyclic subgroup $langle hNrangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N cong H /(H cap N)$, but we have already seen that $H cap N = {e}$, so $HN/N cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
The point is that since $H$ is not a subgroup of $N$, in fact $H cap N = {e}$, since if $h^r in N$ for some $r$, then unless $r equiv 0 mod p$ it has a multiplicative inverse modulo $p$ say $rr' equiv 1 mod p$, in which case $(h^r)^{r'} = h in N$, so $H$ will be a subgroup of $N$,a contradiction.
Now, we may proceed in this way : consider the cyclic group $langle hNrangle subset G/N$. We note that it has at most $p$ elements since $h^pN = N$ as $h^p = e$. However, if $h^rN =N$ for some smaller $r$, then this implies that $h^r in N$, which is a contradiction as we have $h^r in H$, and showed earlier that $H cap N = {e}$. So, the cyclic subgroup $langle hNrangle$ of $G /N$ is isomorphic to $H$, since any two groups of the same prime order are isomorphic.
Alternately, use the second isomorphism theorem : note that $N$ is normal in $G$, and therefore $NH=HN$ is a subgroup of $G$. The second isomorphism theorem asserts that $NH/N cong H /(H cap N)$, but we have already seen that $H cap N = {e}$, so $HN/N cong H$. So one deduces that $HN/N$ is a subgroup of $G/N$ which is isomorphic to $H$.
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 28 '18 at 12:14
астон вілла олоф мэллбэрг
37.3k33376
37.3k33376
add a comment |
add a comment |
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Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53