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






abstract-algebra group-theory







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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






abstract-algebra group-theory







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






abstract-algebra group-theory







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




abstract-algebra group-theory






share|cite|improve this question













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asked Nov 28 '18 at 11:51









John Cataldo

1,0161216




1,0161216








  • 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
4






active

oldest

votes


















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.






      share|cite|improve this answer





























        0














        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





















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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

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          active

          oldest

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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














            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 $H$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      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.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 28 '18 at 12:11









                              Thomas Shelby

                              1,667216




                              1,667216























                                  0














                                  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


























                                    0














                                    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
























                                      0












                                      0








                                      0






                                      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 $H$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 28 '18 at 12:14









                                      астон вілла олоф мэллбэрг

                                      37.3k33376




                                      37.3k33376






























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