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
1,0161216
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
1,0161216
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
1,0161216
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
1,0161216
asked Nov 28 '18 at 11:51
John Cataldo
1,0161216
asked Nov 28 '18 at 11:51
John Cataldo
1,0161216
asked Nov 28 '18 at 11:51
John Cataldo
1,0161216
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
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
1,667216
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017055%2fproving-that-there-exists-a-subgroup-of-g-n-thats-isomorphic-to-h-le-g-te%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
1,667216
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
1,667216
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
1,667216
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
1,667216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,667216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,667216
answered Nov 28 '18 at 12:11
Thomas Shelby
1,667216
1,667216
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017055%2fproving-that-there-exists-a-subgroup-of-g-n-thats-isomorphic-to-h-le-g-te%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Look at the subgroup generated by $hN$.
– Arnaud D.
Nov 28 '18 at 11:53