Proving that there exists a subgroup of $G/N$ that's isomorphic to $Hle G, text{with} |H|=p$ and $Hnleq N$












1














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










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  • 1




    Look at the subgroup generated by $hN$.
    – Arnaud D.
    Nov 28 '18 at 11:53
















1














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










share|cite|improve this question


















  • 1




    Look at the subgroup generated by $hN$.
    – Arnaud D.
    Nov 28 '18 at 11:53














1












1








1







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










share|cite|improve this question













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






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asked Nov 28 '18 at 11:51









John Cataldo

1,0961216




1,0961216








  • 1




    Look at the subgroup generated by $hN$.
    – Arnaud D.
    Nov 28 '18 at 11:53














  • 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










4 Answers
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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 $H$.






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    1














    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






    share|cite|improve this answer





























      1














      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 $H$.






      share|cite|improve this answer





























        0














        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.






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          4 Answers
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          4 Answers
          4






          active

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          active

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          active

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          1














          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 $H$.






          share|cite|improve this answer


























            1














            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 $H$.






            share|cite|improve this answer
























              1












              1








              1






              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 $H$.






              share|cite|improve this answer












              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 $H$.







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              answered Nov 28 '18 at 12:14









              астон вілла олоф мэллбэрг

              37.4k33376




              37.4k33376























                  1














                  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






                  share|cite|improve this answer


























                    1














                    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






                    share|cite|improve this answer
























                      1












                      1








                      1






                      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






                      share|cite|improve this answer












                      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







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 28 '18 at 12:10









                      Shubham

                      1,5951519




                      1,5951519























                          1














                          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 $H$.






                          share|cite|improve this answer


























                            1














                            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 $H$.






                            share|cite|improve this answer
























                              1












                              1








                              1






                              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 $H$.






                              share|cite|improve this answer












                              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 $H$.







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                              answered Nov 28 '18 at 12:12









                              Hebe

                              30818




                              30818























                                  0














                                  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.






                                  share|cite|improve this answer


























                                    0














                                    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.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      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.






                                      share|cite|improve this answer












                                      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.







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                                      answered Nov 28 '18 at 12:11









                                      Thomas Shelby

                                      1,887217




                                      1,887217






























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