Mixing
Prove $KNcong Ktimes N$ if $Kcap N={e}$
$begingroup$
Let $K$, $N$ be normal subgroups in $G$. I would like to show that if $Kcap N={e}$, then $KNcong Ktimes N$.
My original thought was to invoke the first Isomorphism theorem. Therefore I would have to find an appropriate mapping $varphi$ whose kernel is ${e}$.
In a previous exercise, I showed, again by the first Isomorphism theorem that
$$frac{KN}{Kcap N}congfrac{KN}{N}timesfrac{KN}{K}$$
thus, if $Kcap N={e}, $ i would just have to show that $frac{KN}{N}=K$ and $frac{KN}{K}=N$. Thus my idea was to prove that under the condition that the intersection of $K$ and $N$ that elements in $KN$ can commute freely. We know that $KN=NK$, but this does not imply commutativity, only that if $knin KN$, then $kn=n'k'$ with $n$ not necessarily equal to $n'$ and such with $k$. I just can't come up with a commutativity proof.
EDIT: I forgot to define $KN={kn|kin K, nin N}$.
abstract-algebra group-theory
edited Jan 8 at 1:39
the_fox
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
$endgroup$
add a comment |
$begingroup$
Let $K$, $N$ be normal subgroups in $G$. I would like to show that if $Kcap N={e}$, then $KNcong Ktimes N$.
My original thought was to invoke the first Isomorphism theorem. Therefore I would have to find an appropriate mapping $varphi$ whose kernel is ${e}$.
In a previous exercise, I showed, again by the first Isomorphism theorem that
$$frac{KN}{Kcap N}congfrac{KN}{N}timesfrac{KN}{K}$$
thus, if $Kcap N={e}, $ i would just have to show that $frac{KN}{N}=K$ and $frac{KN}{K}=N$. Thus my idea was to prove that under the condition that the intersection of $K$ and $N$ that elements in $KN$ can commute freely. We know that $KN=NK$, but this does not imply commutativity, only that if $knin KN$, then $kn=n'k'$ with $n$ not necessarily equal to $n'$ and such with $k$. I just can't come up with a commutativity proof.
EDIT: I forgot to define $KN={kn|kin K, nin N}$.
abstract-algebra group-theory
edited Jan 8 at 1:39
the_fox
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
$endgroup$
2
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36
add a comment |
$begingroup$
Let $K$, $N$ be normal subgroups in $G$. I would like to show that if $Kcap N={e}$, then $KNcong Ktimes N$.
My original thought was to invoke the first Isomorphism theorem. Therefore I would have to find an appropriate mapping $varphi$ whose kernel is ${e}$.
In a previous exercise, I showed, again by the first Isomorphism theorem that
$$frac{KN}{Kcap N}congfrac{KN}{N}timesfrac{KN}{K}$$
thus, if $Kcap N={e}, $ i would just have to show that $frac{KN}{N}=K$ and $frac{KN}{K}=N$. Thus my idea was to prove that under the condition that the intersection of $K$ and $N$ that elements in $KN$ can commute freely. We know that $KN=NK$, but this does not imply commutativity, only that if $knin KN$, then $kn=n'k'$ with $n$ not necessarily equal to $n'$ and such with $k$. I just can't come up with a commutativity proof.
EDIT: I forgot to define $KN={kn|kin K, nin N}$.
abstract-algebra group-theory
edited Jan 8 at 1:39
the_fox
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
$endgroup$
Let $K$, $N$ be normal subgroups in $G$. I would like to show that if $Kcap N={e}$, then $KNcong Ktimes N$.
My original thought was to invoke the first Isomorphism theorem. Therefore I would have to find an appropriate mapping $varphi$ whose kernel is ${e}$.
In a previous exercise, I showed, again by the first Isomorphism theorem that
$$frac{KN}{Kcap N}congfrac{KN}{N}timesfrac{KN}{K}$$
thus, if $Kcap N={e}, $ i would just have to show that $frac{KN}{N}=K$ and $frac{KN}{K}=N$. Thus my idea was to prove that under the condition that the intersection of $K$ and $N$ that elements in $KN$ can commute freely. We know that $KN=NK$, but this does not imply commutativity, only that if $knin KN$, then $kn=n'k'$ with $n$ not necessarily equal to $n'$ and such with $k$. I just can't come up with a commutativity proof.
EDIT: I forgot to define $KN={kn|kin K, nin N}$.
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 8 at 1:39
the_fox
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
edited Jan 8 at 1:39
the_fox
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
edited Jan 8 at 1:39
the_fox
2,64221534
edited Jan 8 at 1:39
the_fox
2,64221534
edited Jan 8 at 1:39
the_fox
2,64221534
2,64221534
asked Dec 16 '15 at 1:30
IcemanIceman
749721
asked Dec 16 '15 at 1:30
IcemanIceman
749721
asked Dec 16 '15 at 1:30
IcemanIceman
749721
749721
2
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36
add a comment |
2
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36
2
2
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To show that $KN / N cong K$, use an isomorphism theorem. Take the map
$$varphi : KN to K, quad kn to k$$
To check that this is well-defined, notice that
$$kn = k'n' iff k'^{-1}k = n'n^{-1} iff k = k' text{ and } n' = n$$
It's clearly a homomorphism, and its kernel is exactly $N$.
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
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%2f1577720%2fprove-kn-cong-k-times-n-if-k-cap-n-e%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To show that $KN / N cong K$, use an isomorphism theorem. Take the map
$$varphi : KN to K, quad kn to k$$
To check that this is well-defined, notice that
$$kn = k'n' iff k'^{-1}k = n'n^{-1} iff k = k' text{ and } n' = n$$
It's clearly a homomorphism, and its kernel is exactly $N$.
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
add a comment |
$begingroup$
To show that $KN / N cong K$, use an isomorphism theorem. Take the map
$$varphi : KN to K, quad kn to k$$
To check that this is well-defined, notice that
$$kn = k'n' iff k'^{-1}k = n'n^{-1} iff k = k' text{ and } n' = n$$
It's clearly a homomorphism, and its kernel is exactly $N$.
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
add a comment |
$begingroup$
To show that $KN / N cong K$, use an isomorphism theorem. Take the map
$$varphi : KN to K, quad kn to k$$
To check that this is well-defined, notice that
$$kn = k'n' iff k'^{-1}k = n'n^{-1} iff k = k' text{ and } n' = n$$
It's clearly a homomorphism, and its kernel is exactly $N$.
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
$endgroup$
To show that $KN / N cong K$, use an isomorphism theorem. Take the map
$$varphi : KN to K, quad kn to k$$
To check that this is well-defined, notice that
$$kn = k'n' iff k'^{-1}k = n'n^{-1} iff k = k' text{ and } n' = n$$
It's clearly a homomorphism, and its kernel is exactly $N$.
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
answered Dec 16 '15 at 1:34
T. BongersT. Bongers
23.1k54662
23.1k54662
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
add a comment |
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
$begingroup$
I am having trouble with the homomorphism, since you need two elements from $KN$, $k_1n_1, k_2n_2$. So $varphi(k_1n_1k_2n_2)=varphi(k_1k_2'n_1'n_2)=k_1k_2'=...$
$endgroup$
– Iceman
Dec 16 '15 at 1:54
1
1
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
$begingroup$
@Iceman Compute the commutator of two elements and you'll find that $kn = nk$ for all $n, k$. For example, $kn(nk)^{-1} = knk^{-1} cdot n^{-1} in N$, while $kn(nk)^{-1} = k cdot nk^{-1}n^{-1} in K$.
$endgroup$
– T. Bongers
Dec 16 '15 at 2:27
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.
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%2f1577720%2fprove-kn-cong-k-times-n-if-k-cap-n-e%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
2
$begingroup$
The answer below is perfect for the tools you have at this level, but this also follows directly from the semidirect product.
$endgroup$
– Alan
Dec 16 '15 at 1:36