Image of a subgroup is subgroup under homomorphism. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)the image of a normal subgroupGroup Homomorphism: Kernel, Image, quotient groupIs image of homomorphism is a subgroup?The preimage of a normal subgroup under a group homomorphism is normalUsing a homomorphism to draw conclusions about subgroup indicesLet G be simple, |G| is not 2, and ϕ a homomorphism from G to H. If H contains a normal subgroup A of index 2, then ϕ(G) ≤ A .Ring homomorphism: Prove the image is a subringImage of a subgroup under a homomorphism is not a normal subgroup?What does normality is image closed mean here?Is every normal subgroup the kernel of some self-homomorphism?
An adverb for when you're not exaggerating
Should I use a zero-interest credit card for a large one-time purchase?
Is grep documentation wrong?
What is homebrew?
Dating a Former Employee
Using et al. for a last / senior author rather than for a first author
Is this homebrew Lady of Pain warlock patron balanced?
Maximum summed powersets with non-adjacent items
When the Haste spell ends on a creature, do attackers have advantage against that creature?
Do wooden building fires get hotter than 600°C?
Do square wave exist?
Can an alien society believe that their star system is the universe?
Why are the trig functions versine, haversine, exsecant, etc, rarely used in modern mathematics?
2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?
If a VARCHAR(MAX) column is included in an index, is the entire value always stored in the index page(s)?
Can anything be seen from the center of the Boötes void? How dark would it be?
Irreducible of finite Krull dimension implies quasi-compact?
Where are Serre’s lectures at Collège de France to be found?
Do I really need recursive chmod to restrict access to a folder?
What is the meaning of the new sigil in Game of Thrones Season 8 intro?
How to convince students of the implication truth values?
Can a new player join a group only when a new campaign starts?
Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?
Why do we bend a book to keep it straight?
Image of a subgroup is subgroup under homomorphism.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)the image of a normal subgroupGroup Homomorphism: Kernel, Image, quotient groupIs image of homomorphism is a subgroup?The preimage of a normal subgroup under a group homomorphism is normalUsing a homomorphism to draw conclusions about subgroup indicesLet G be simple, |G| is not 2, and ϕ a homomorphism from G to H. If H contains a normal subgroup A of index 2, then ϕ(G) ≤ A .Ring homomorphism: Prove the image is a subringImage of a subgroup under a homomorphism is not a normal subgroup?What does normality is image closed mean here?Is every normal subgroup the kernel of some self-homomorphism?
$begingroup$
Image of a subgroup under group homomorphism is a subgroup of "codomain."
Image of a normal subgroup under group homomorphism is a normal subgroup of "range."
Image of a subring under ring homomorphism is a subring of "codomain."
Did I state correctly?
I've been confused about codomain and range.
abstract-algebra normal-subgroups group-homomorphism ring-homomorphism
asked Mar 27 at 5:28
SophiaSophia
7717
$endgroup$
add a comment |
$begingroup$
Image of a subgroup under group homomorphism is a subgroup of "codomain."
Image of a normal subgroup under group homomorphism is a normal subgroup of "range."
Image of a subring under ring homomorphism is a subring of "codomain."
Did I state correctly?
I've been confused about codomain and range.
abstract-algebra normal-subgroups group-homomorphism ring-homomorphism
asked Mar 27 at 5:28
SophiaSophia
7717
$endgroup$
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29
add a comment |
$begingroup$
Image of a subgroup under group homomorphism is a subgroup of "codomain."
Image of a normal subgroup under group homomorphism is a normal subgroup of "range."
Image of a subring under ring homomorphism is a subring of "codomain."
Did I state correctly?
I've been confused about codomain and range.
abstract-algebra normal-subgroups group-homomorphism ring-homomorphism
asked Mar 27 at 5:28
SophiaSophia
7717
$endgroup$
Image of a subgroup under group homomorphism is a subgroup of "codomain."
Image of a normal subgroup under group homomorphism is a normal subgroup of "range."
Image of a subring under ring homomorphism is a subring of "codomain."
Did I state correctly?
I've been confused about codomain and range.
abstract-algebra normal-subgroups group-homomorphism ring-homomorphism
abstract-algebra normal-subgroups group-homomorphism ring-homomorphism
asked Mar 27 at 5:28
SophiaSophia
7717
asked Mar 27 at 5:28
SophiaSophia
7717
asked Mar 27 at 5:28
SophiaSophia
7717
asked Mar 27 at 5:28
SophiaSophia
7717
asked Mar 27 at 5:28
SophiaSophia
7717
7717
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29
add a comment |
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29
add a comment |
0
active
oldest
votes
Your Answer
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%2f3164130%2fimage-of-a-subgroup-is-subgroup-under-homomorphism%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3164130%2fimage-of-a-subgroup-is-subgroup-under-homomorphism%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
$begingroup$
This is all correct but there's a slightly easier way to think about this by reframing some of these. If you know that, in general, the image of any group under a homomorphism is a subgroup of the codomain, think about the image of any subgroup of that group as just the image of the same homomorphism but with a restricted domain. The distinction for codomain and range kicks in for normalcy because it relies on the fact that each element in the range is the image of some codomain element so if $u,vin R=f(N)$, $uvu^-1=f(x)f(y)f(x)^-1, x,yin N$ $=f(xyx^-1)=f(r), rin N$ $=w, win f(N)$
$endgroup$
– Cardioid_Ass_22
Mar 27 at 6:29