When is it true that $nexists gne e$ such that $phi(g)=e$ for a group homomorphism $phi: Gmapsto H$ and...
I was thinking if it is possible to have $gne e$ such that $phi(g)=e$ for a group homomorphism $phi: Gmapsto H$
It's not always true because $phi(2)=0$ when $G=Bbb Z_4$ and $H=Bbb Z_2$
So what if $|H|notmid|G|$?
We have that $o(phi(g))$ divides $o(g)$ and $|H|$ because $phi(langle grangle)$ is a subgroup of $H$
If we had $(|H|,phi(langle grangle))=1$ then we would have $o(phi(g))=1$ and so $phi(g)=e$
What other necessary/sufficient constraints are there for the existance of $gin Gbackslash{e}$ with $phi(g)=e$?
abstract-algebra group-theory group-homomorphism
edited Jan 4 at 13:41
Scientifica
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
add a comment |
I was thinking if it is possible to have $gne e$ such that $phi(g)=e$ for a group homomorphism $phi: Gmapsto H$
It's not always true because $phi(2)=0$ when $G=Bbb Z_4$ and $H=Bbb Z_2$
So what if $|H|notmid|G|$?
We have that $o(phi(g))$ divides $o(g)$ and $|H|$ because $phi(langle grangle)$ is a subgroup of $H$
If we had $(|H|,phi(langle grangle))=1$ then we would have $o(phi(g))=1$ and so $phi(g)=e$
What other necessary/sufficient constraints are there for the existance of $gin Gbackslash{e}$ with $phi(g)=e$?
abstract-algebra group-theory group-homomorphism
edited Jan 4 at 13:41
Scientifica
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
1
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
4
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25
add a comment |
I was thinking if it is possible to have $gne e$ such that $phi(g)=e$ for a group homomorphism $phi: Gmapsto H$
It's not always true because $phi(2)=0$ when $G=Bbb Z_4$ and $H=Bbb Z_2$
So what if $|H|notmid|G|$?
We have that $o(phi(g))$ divides $o(g)$ and $|H|$ because $phi(langle grangle)$ is a subgroup of $H$
If we had $(|H|,phi(langle grangle))=1$ then we would have $o(phi(g))=1$ and so $phi(g)=e$
What other necessary/sufficient constraints are there for the existance of $gin Gbackslash{e}$ with $phi(g)=e$?
abstract-algebra group-theory group-homomorphism
edited Jan 4 at 13:41
Scientifica
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
I was thinking if it is possible to have $gne e$ such that $phi(g)=e$ for a group homomorphism $phi: Gmapsto H$
It's not always true because $phi(2)=0$ when $G=Bbb Z_4$ and $H=Bbb Z_2$
So what if $|H|notmid|G|$?
We have that $o(phi(g))$ divides $o(g)$ and $|H|$ because $phi(langle grangle)$ is a subgroup of $H$
If we had $(|H|,phi(langle grangle))=1$ then we would have $o(phi(g))=1$ and so $phi(g)=e$
What other necessary/sufficient constraints are there for the existance of $gin Gbackslash{e}$ with $phi(g)=e$?
abstract-algebra group-theory group-homomorphism
abstract-algebra group-theory group-homomorphism
edited Jan 4 at 13:41
Scientifica
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
edited Jan 4 at 13:41
Scientifica
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
edited Jan 4 at 13:41
Scientifica
6,37641335
edited Jan 4 at 13:41
Scientifica
6,37641335
edited Jan 4 at 13:41
Scientifica
6,37641335
6,37641335
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
asked Jan 4 at 13:16
John CataldoJohn Cataldo
1,0961216
1,0961216
1
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
4
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25
add a comment |
1
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
4
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25
1
1
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
4
4
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25
add a comment |
1 Answer
1
active
oldest
votes
First of all, let's use a fact to make your question a bit neater.
The following conditions are equivalent:
$phi : G to H$ is injective (one-to-one)
- $ker phi = {e}$
- There does not exist an element $g in G$ with $g neq e$ such that $phi(g) = e$
With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = Bbb Z_2$ and $H = Bbb Z_4$, we see that the map $phi([n]_2) = [2n]_4$ is injective, and $Bbb Z_4$ contains the subgroup ${[0]_4,[2]_4}$ which is isomorphic to $Bbb Z_2$.
One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $phi(G) cong G/kerphi$. However, if $ker phi = {e}$, then we have $G/ker phi cong G$. So, if $phi$ is an injective homomorphism, then $phi(G) cong G$, and $phi(G)$ (the image of $phi$) is a subgroup of $H$.
Conversely, suppose that $H$ has a subgroup $K subset H$ and that $K cong G$. Then, an isomorphism $phi:G to K$ means that we have the injective homomorphism $i_K circ phi:G to H$, where $i_K:K to H$ is the inclusion map.
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
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%2f3061639%2fwhen-is-it-true-that-nexists-g-ne-e-such-that-phig-e-for-a-group-homomor%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
First of all, let's use a fact to make your question a bit neater.
The following conditions are equivalent:
$phi : G to H$ is injective (one-to-one)
- $ker phi = {e}$
- There does not exist an element $g in G$ with $g neq e$ such that $phi(g) = e$
With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = Bbb Z_2$ and $H = Bbb Z_4$, we see that the map $phi([n]_2) = [2n]_4$ is injective, and $Bbb Z_4$ contains the subgroup ${[0]_4,[2]_4}$ which is isomorphic to $Bbb Z_2$.
One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $phi(G) cong G/kerphi$. However, if $ker phi = {e}$, then we have $G/ker phi cong G$. So, if $phi$ is an injective homomorphism, then $phi(G) cong G$, and $phi(G)$ (the image of $phi$) is a subgroup of $H$.
Conversely, suppose that $H$ has a subgroup $K subset H$ and that $K cong G$. Then, an isomorphism $phi:G to K$ means that we have the injective homomorphism $i_K circ phi:G to H$, where $i_K:K to H$ is the inclusion map.
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
add a comment |
First of all, let's use a fact to make your question a bit neater.
The following conditions are equivalent:
$phi : G to H$ is injective (one-to-one)
- $ker phi = {e}$
- There does not exist an element $g in G$ with $g neq e$ such that $phi(g) = e$
With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = Bbb Z_2$ and $H = Bbb Z_4$, we see that the map $phi([n]_2) = [2n]_4$ is injective, and $Bbb Z_4$ contains the subgroup ${[0]_4,[2]_4}$ which is isomorphic to $Bbb Z_2$.
One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $phi(G) cong G/kerphi$. However, if $ker phi = {e}$, then we have $G/ker phi cong G$. So, if $phi$ is an injective homomorphism, then $phi(G) cong G$, and $phi(G)$ (the image of $phi$) is a subgroup of $H$.
Conversely, suppose that $H$ has a subgroup $K subset H$ and that $K cong G$. Then, an isomorphism $phi:G to K$ means that we have the injective homomorphism $i_K circ phi:G to H$, where $i_K:K to H$ is the inclusion map.
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
add a comment |
First of all, let's use a fact to make your question a bit neater.
The following conditions are equivalent:
$phi : G to H$ is injective (one-to-one)
- $ker phi = {e}$
- There does not exist an element $g in G$ with $g neq e$ such that $phi(g) = e$
With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = Bbb Z_2$ and $H = Bbb Z_4$, we see that the map $phi([n]_2) = [2n]_4$ is injective, and $Bbb Z_4$ contains the subgroup ${[0]_4,[2]_4}$ which is isomorphic to $Bbb Z_2$.
One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $phi(G) cong G/kerphi$. However, if $ker phi = {e}$, then we have $G/ker phi cong G$. So, if $phi$ is an injective homomorphism, then $phi(G) cong G$, and $phi(G)$ (the image of $phi$) is a subgroup of $H$.
Conversely, suppose that $H$ has a subgroup $K subset H$ and that $K cong G$. Then, an isomorphism $phi:G to K$ means that we have the injective homomorphism $i_K circ phi:G to H$, where $i_K:K to H$ is the inclusion map.
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
First of all, let's use a fact to make your question a bit neater.
The following conditions are equivalent:
$phi : G to H$ is injective (one-to-one)
- $ker phi = {e}$
- There does not exist an element $g in G$ with $g neq e$ such that $phi(g) = e$
With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = Bbb Z_2$ and $H = Bbb Z_4$, we see that the map $phi([n]_2) = [2n]_4$ is injective, and $Bbb Z_4$ contains the subgroup ${[0]_4,[2]_4}$ which is isomorphic to $Bbb Z_2$.
One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $phi(G) cong G/kerphi$. However, if $ker phi = {e}$, then we have $G/ker phi cong G$. So, if $phi$ is an injective homomorphism, then $phi(G) cong G$, and $phi(G)$ (the image of $phi$) is a subgroup of $H$.
Conversely, suppose that $H$ has a subgroup $K subset H$ and that $K cong G$. Then, an isomorphism $phi:G to K$ means that we have the injective homomorphism $i_K circ phi:G to H$, where $i_K:K to H$ is the inclusion map.
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
edited Jan 4 at 13:52
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
answered Jan 4 at 13:39
OmnomnomnomOmnomnomnom
127k788176
127k788176
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
add a comment |
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
@Mark good catch, thanks
– Omnomnomnom
Jan 4 at 13:42
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
Thanks! (I think you meant big $G$ and $Kcong G$)
– John Cataldo
Jan 4 at 13:49
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
@JohnCataldo I don't see what you mean by "big $G$"
– Omnomnomnom
Jan 4 at 13:50
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
"$H$ contains a subgroup isomorphic to $g$"
– John Cataldo
Jan 4 at 13:51
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
@JohnCataldo that makes sense now, thanks
– Omnomnomnom
Jan 4 at 13:53
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%2f3061639%2fwhen-is-it-true-that-nexists-g-ne-e-such-that-phig-e-for-a-group-homomor%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
Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $Gto H$ which takes every element to the identity.
– lulu
Jan 4 at 13:18
My question is about the conditions that would assure $exists gin G~~:~phi(gne e)=e$. So $phi=id$ is a condition but my question is more general
– John Cataldo
Jan 4 at 13:20
Well, if $phi$ has no non-trivial elements in its kernel then $phi(G)$ is isomorphic to $G$, in which case of course we have $|H|, |,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$.
– lulu
Jan 4 at 13:22
4
A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$.
– Omnomnomnom
Jan 4 at 13:22
And indeed, if $|G| nmid |H|$, then there are no injective homomorphisms from $G$ to $H$.
– Omnomnomnom
Jan 4 at 13:25