Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$. [closed]
$begingroup$
Let $G$ be a finite group, $Nmathrel{lhd}G$ a normal subgroup of $G$, and $Hleq G$ a subgroup of $G$. Suppose that $|H|$ and $|N|$ are relatively prime (i.e., $gcd(|H|,|N|)=1$). Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$.
I am trying to show that there is an injective homomorphism $varphi:Hrightarrow G/N$ , but I have no clue for the next step.
group-theory finite-groups normal-subgroups group-isomorphism quotient-group
edited Dec 8 '18 at 23:13
Batominovski
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
$endgroup$
closed as unclear what you're asking by Shaun, user10354138, Cesareo, Rebellos, GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 14:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $G$ be a finite group, $Nmathrel{lhd}G$ a normal subgroup of $G$, and $Hleq G$ a subgroup of $G$. Suppose that $|H|$ and $|N|$ are relatively prime (i.e., $gcd(|H|,|N|)=1$). Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$.
I am trying to show that there is an injective homomorphism $varphi:Hrightarrow G/N$ , but I have no clue for the next step.
group-theory finite-groups normal-subgroups group-isomorphism quotient-group
edited Dec 8 '18 at 23:13
Batominovski
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
$endgroup$
closed as unclear what you're asking by Shaun, user10354138, Cesareo, Rebellos, GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 14:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
1
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16
add a comment |
$begingroup$
Let $G$ be a finite group, $Nmathrel{lhd}G$ a normal subgroup of $G$, and $Hleq G$ a subgroup of $G$. Suppose that $|H|$ and $|N|$ are relatively prime (i.e., $gcd(|H|,|N|)=1$). Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$.
I am trying to show that there is an injective homomorphism $varphi:Hrightarrow G/N$ , but I have no clue for the next step.
group-theory finite-groups normal-subgroups group-isomorphism quotient-group
edited Dec 8 '18 at 23:13
Batominovski
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
$endgroup$
Let $G$ be a finite group, $Nmathrel{lhd}G$ a normal subgroup of $G$, and $Hleq G$ a subgroup of $G$. Suppose that $|H|$ and $|N|$ are relatively prime (i.e., $gcd(|H|,|N|)=1$). Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$.
I am trying to show that there is an injective homomorphism $varphi:Hrightarrow G/N$ , but I have no clue for the next step.
group-theory finite-groups normal-subgroups group-isomorphism quotient-group
group-theory finite-groups normal-subgroups group-isomorphism quotient-group
edited Dec 8 '18 at 23:13
Batominovski
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
edited Dec 8 '18 at 23:13
Batominovski
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
edited Dec 8 '18 at 23:13
Batominovski
33k33293
edited Dec 8 '18 at 23:13
Batominovski
33k33293
edited Dec 8 '18 at 23:13
Batominovski
33k33293
33k33293
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
asked Dec 8 '18 at 22:29
aaron wangaaron wang
162
162
closed as unclear what you're asking by Shaun, user10354138, Cesareo, Rebellos, GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 14:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Shaun, user10354138, Cesareo, Rebellos, GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 14:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
1
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16
add a comment |
4
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
1
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16
4
4
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
1
1
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Show that $HN$ is a subgroup of $G$ containing $N$. Prove that, under the canonical projection $Gto (G/N)$, the subgroup $HN$ of $G$ is mapped onto a subgroup of $G/N$ isomorphic to $H$ (this subgroup is clearly, $HN/N$). More generally, we have $HN/Ncong H/(Hcap N)$.
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Show that $HN$ is a subgroup of $G$ containing $N$. Prove that, under the canonical projection $Gto (G/N)$, the subgroup $HN$ of $G$ is mapped onto a subgroup of $G/N$ isomorphic to $H$ (this subgroup is clearly, $HN/N$). More generally, we have $HN/Ncong H/(Hcap N)$.
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
$endgroup$
add a comment |
$begingroup$
Hint: Show that $HN$ is a subgroup of $G$ containing $N$. Prove that, under the canonical projection $Gto (G/N)$, the subgroup $HN$ of $G$ is mapped onto a subgroup of $G/N$ isomorphic to $H$ (this subgroup is clearly, $HN/N$). More generally, we have $HN/Ncong H/(Hcap N)$.
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
$endgroup$
add a comment |
$begingroup$
Hint: Show that $HN$ is a subgroup of $G$ containing $N$. Prove that, under the canonical projection $Gto (G/N)$, the subgroup $HN$ of $G$ is mapped onto a subgroup of $G/N$ isomorphic to $H$ (this subgroup is clearly, $HN/N$). More generally, we have $HN/Ncong H/(Hcap N)$.
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
$endgroup$
Hint: Show that $HN$ is a subgroup of $G$ containing $N$. Prove that, under the canonical projection $Gto (G/N)$, the subgroup $HN$ of $G$ is mapped onto a subgroup of $G/N$ isomorphic to $H$ (this subgroup is clearly, $HN/N$). More generally, we have $HN/Ncong H/(Hcap N)$.
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
answered Dec 8 '18 at 23:11
BatominovskiBatominovski
33k33293
33k33293
add a comment |
add a comment |
4
$begingroup$
Do You mean $N triangleleft G$ instead of $H triangleleft G$? And also $G/N$ instead of $G/H$? as $G/H$ may not be a group.
$endgroup$
– nafhgood
Dec 8 '18 at 22:30
$begingroup$
Hint. Think about what the natural homomorphism from $G$ to $G/N$ does to $H$. Can it send anything other than the identity to the identity?
$endgroup$
– Ethan Bolker
Dec 8 '18 at 23:03
1
$begingroup$
The second isomorphism theorem says $HN/N cong H/(H cap N)$. Here $H cap N$ is trivial as $(|H|,|N|)=1$. So then $HN/N cong H$.
$endgroup$
– nafhgood
Dec 8 '18 at 23:16