Few group theory questions
I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$
I can show that it is reflexive as the identity is always in the subgroup.
if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all
group-theory relations group-homomorphism
edited Nov 28 '18 at 22:43
the_fox
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
add a comment |
I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$
I can show that it is reflexive as the identity is always in the subgroup.
if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all
group-theory relations group-homomorphism
edited Nov 28 '18 at 22:43
the_fox
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32
add a comment |
I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$
I can show that it is reflexive as the identity is always in the subgroup.
if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all
group-theory relations group-homomorphism
edited Nov 28 '18 at 22:43
the_fox
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a cong b$ if $b^{-1}a in H$
I can show that it is reflexive as the identity is always in the subgroup.
if $a cong b$ then $b^{-1}a in H$ and so $(b^{-1}a)^{-1}=a^{-1}b in H$ so $b cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a cong b $ then $b^{-1}a in H $ and if $b cong c$ then $c^{-1}b in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $mathbb{Z}/nmathbb{Z}$ to $mathbb{Z}$ and it says for $n ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $phi$ was a homomorphism then $phi(ab)=phi(a)phi(b)$ and this would quickly result in having $ab=n=0$ but $phi(a)$ and $phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all
group-theory relations group-homomorphism
group-theory relations group-homomorphism
edited Nov 28 '18 at 22:43
the_fox
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
edited Nov 28 '18 at 22:43
the_fox
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
edited Nov 28 '18 at 22:43
the_fox
2,50711431
edited Nov 28 '18 at 22:43
the_fox
2,50711431
edited Nov 28 '18 at 22:43
the_fox
2,50711431
2,50711431
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
asked Dec 8 '15 at 21:17
PersonaAPersonaA
329617
329617
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32
add a comment |
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32
add a comment |
2 Answers
2
active
oldest
votes
Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.
So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.
Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$
And $an = 0$ for all $a$ in $Bbb Z/n$.
So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.
So $phi$ is the zero homomorphism. It's the only possible homomorphism.
edited Nov 28 '18 at 22:21
Shaun
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
add a comment |
- Correct, commutativity is not needed and not used in your proof.
- The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
|
show 1 more 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%2f1566521%2ffew-group-theory-questions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.
So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.
Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$
And $an = 0$ for all $a$ in $Bbb Z/n$.
So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.
So $phi$ is the zero homomorphism. It's the only possible homomorphism.
edited Nov 28 '18 at 22:21
Shaun
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
add a comment |
Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.
So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.
Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$
And $an = 0$ for all $a$ in $Bbb Z/n$.
So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.
So $phi$ is the zero homomorphism. It's the only possible homomorphism.
edited Nov 28 '18 at 22:21
Shaun
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
add a comment |
Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.
So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.
Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$
And $an = 0$ for all $a$ in $Bbb Z/n$.
So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.
So $phi$ is the zero homomorphism. It's the only possible homomorphism.
edited Nov 28 '18 at 22:21
Shaun
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
Let $phi:Bbb Z/nto Bbb Z$ be a homomorphism.
So $phi(a + b) = phi(a) + phi(b)$ so $phi(a) = phi(0 + a) = phi(0) + phi(a)$ so $phi(0) = 0$.
Now $$begin{align}phi(am) &= phileft(underbrace{a + a + dots + a}_{mtext{ times.}}right) \ &= underbrace{phi(a) + phi(a) + dots + phi(a)}_{mtext{ times.}} \ &= phi(a)times m.end{align}$$
And $an = 0$ for all $a$ in $Bbb Z/n$.
So $phi(an) = phi(a)times n = 0$ for all $a$ in $Bbb Z/n$. But in $phi(a) in Bbb Z$. $phi(a)times n = 0 implies phi(a) = 0$. This is true for all $a$ in $Bbb Z/n$.
So $phi$ is the zero homomorphism. It's the only possible homomorphism.
edited Nov 28 '18 at 22:21
Shaun
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
edited Nov 28 '18 at 22:21
Shaun
8,820113681
edited Nov 28 '18 at 22:21
Shaun
8,820113681
edited Nov 28 '18 at 22:21
Shaun
8,820113681
8,820113681
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
answered Dec 8 '15 at 21:53
fleabloodfleablood
68.6k22685
68.6k22685
add a comment |
add a comment |
- Correct, commutativity is not needed and not used in your proof.
- The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
|
show 1 more comment
- Correct, commutativity is not needed and not used in your proof.
- The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
|
show 1 more comment
- Correct, commutativity is not needed and not used in your proof.
- The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
- Correct, commutativity is not needed and not used in your proof.
- The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{nBbb Z}$ be mapped by a homomorphism to $Bbb Z$?
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
answered Dec 8 '15 at 21:23
BerciBerci
59.8k23672
59.8k23672
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
|
show 1 more comment
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no?
– PersonaA
Dec 8 '15 at 21:35
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $in$ Z have the property an = 0?
– fleablood
Dec 8 '15 at 21:43
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
Wouldn't it be all of them relatively ptime?
– PersonaA
Dec 8 '15 at 21:47
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n
– fleablood
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
Only a=0 then ?
– PersonaA
Dec 8 '15 at 21:48
|
show 1 more 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%2f1566521%2ffew-group-theory-questions%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
I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0.
– fleablood
Dec 8 '15 at 21:32