Modulo groups and non-prime numbersWhy is group theory axiomatized with operations?Problem in modular arithmetic using group theoryFermat's Little Theorem: group and multiplication moduloExistence of a generator over multiplication for integers modulo pExistence of non-abelian group of order nGroups with Complex NumbersMultiplicatively written and non-abelian groupsSeeing composition tableSuppose $p$ is an odd prime and $G$ is a non-abelian group of order $2p$. Prove that $G$ contains an element of order $p$.Forming a multiplicative group $mod n$ whose order is a prime numberIf we say “classes of non-zero integers modulo $n$”, why does this not include the $0$ class?
What will the Frenchman say?
Why is "la Gestapo" feminine?
Why does Surtur say that Thor is Asgard's doom?
Do I need to convey a moral for each of my blog post?
DisplayForm problem with pi in FractionBox
Why I don't get the wanted width of tcbox?
Why doesn't the chatan sign the ketubah?
Air travel with refrigerated insulin
Imaginary part of expression too difficult to calculate
Why is this tree refusing to shed its dead leaves?
How to find the largest number(s) in a list of elements, possibly non-unique?
Why do I have a large white artefact on the rendered image?
Determine voltage drop over 10G resistors with cheap multimeter
Is this Pascal's Matrix?
pipe commands inside find -exec?
Print last inputted byte
Isn't the word "experience" wrongly used in this context?
Can a university suspend a student even when he has left university?
Jem'Hadar, something strange about their life expectancy
TDE Master Key Rotation
Should I be concerned about student access to a test bank?
What is the reasoning behind standardization (dividing by standard deviation)?
Have the tides ever turned twice on any open problem?
How to test the sharpness of a knife?
Modulo groups and non-prime numbers
Why is group theory axiomatized with operations?Problem in modular arithmetic using group theoryFermat's Little Theorem: group and multiplication moduloExistence of a generator over multiplication for integers modulo pExistence of non-abelian group of order nGroups with Complex NumbersMultiplicatively written and non-abelian groupsSeeing composition tableSuppose $p$ is an odd prime and $G$ is a non-abelian group of order $2p$. Prove that $G$ contains an element of order $p$.Forming a multiplicative group $mod n$ whose order is a prime numberIf we say “classes of non-zero integers modulo $n$”, why does this not include the $0$ class?
$begingroup$
Let $M$ be a non prime number and $G$ be the set of non-zero integers modulo $M$, under multiplication modulo $M$.
Show this is not a group.
My attempt:
Since $M$ is non prime so there exists integers $r,s>1$ such that $rs=M$,
that is $rs=0mod M$, so there exists integers in $G$ where closure doesn't hold. Is this correct?
Also can a left coset equal right coset if group is not abelian?
Thanks in advance.
group-theory
edited Mar 13 at 9:12
user1729
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
$endgroup$
add a comment |
$begingroup$
Let $M$ be a non prime number and $G$ be the set of non-zero integers modulo $M$, under multiplication modulo $M$.
Show this is not a group.
My attempt:
Since $M$ is non prime so there exists integers $r,s>1$ such that $rs=M$,
that is $rs=0mod M$, so there exists integers in $G$ where closure doesn't hold. Is this correct?
Also can a left coset equal right coset if group is not abelian?
Thanks in advance.
group-theory
edited Mar 13 at 9:12
user1729
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
$endgroup$
1
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
1
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
1
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
1
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48
add a comment |
$begingroup$
Let $M$ be a non prime number and $G$ be the set of non-zero integers modulo $M$, under multiplication modulo $M$.
Show this is not a group.
My attempt:
Since $M$ is non prime so there exists integers $r,s>1$ such that $rs=M$,
that is $rs=0mod M$, so there exists integers in $G$ where closure doesn't hold. Is this correct?
Also can a left coset equal right coset if group is not abelian?
Thanks in advance.
group-theory
edited Mar 13 at 9:12
user1729
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
$endgroup$
Let $M$ be a non prime number and $G$ be the set of non-zero integers modulo $M$, under multiplication modulo $M$.
Show this is not a group.
My attempt:
Since $M$ is non prime so there exists integers $r,s>1$ such that $rs=M$,
that is $rs=0mod M$, so there exists integers in $G$ where closure doesn't hold. Is this correct?
Also can a left coset equal right coset if group is not abelian?
Thanks in advance.
group-theory
group-theory
edited Mar 13 at 9:12
user1729
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
edited Mar 13 at 9:12
user1729
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
edited Mar 13 at 9:12
user1729
17.4k64193
edited Mar 13 at 9:12
user1729
17.4k64193
edited Mar 13 at 9:12
user1729
17.4k64193
17.4k64193
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
asked Mar 12 at 15:41
Eden HazardEden Hazard
307
307
1
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
1
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
1
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
1
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48
add a comment |
1
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
1
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
1
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
1
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48
1
1
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
1
1
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
1
1
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
1
1
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, circ)$, must satisfy four requirements known as the group axioms:
Closure. For all $a, bin G$ their product $a circ b$ is an element of $G$.
Associativity. For all $a, b, c in G$ we have $(a circ b) circ c = a circ (b circ c)$.
Identity element. There exists an element $e in G$ such that, for every element $a in G$, the equation $e circ a = a circ e = a$ holds.
Inverse element. For each $a in G$, there exists an element $b in G$ such that $a circ b = b circ a = e$, where $e$ is an (the) identity element.
As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=1,2,3,4,5$ but $2times3=6notin G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)
You also ask:
Can a left coset equal right coset if group is not abelian?
Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=1, (123)(132)$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
answered Mar 13 at 9:06
user1729user1729
17.4k64193
$endgroup$
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
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%2f3145241%2fmodulo-groups-and-non-prime-numbers%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
$begingroup$
There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, circ)$, must satisfy four requirements known as the group axioms:
Closure. For all $a, bin G$ their product $a circ b$ is an element of $G$.
Associativity. For all $a, b, c in G$ we have $(a circ b) circ c = a circ (b circ c)$.
Identity element. There exists an element $e in G$ such that, for every element $a in G$, the equation $e circ a = a circ e = a$ holds.
Inverse element. For each $a in G$, there exists an element $b in G$ such that $a circ b = b circ a = e$, where $e$ is an (the) identity element.
As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=1,2,3,4,5$ but $2times3=6notin G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)
You also ask:
Can a left coset equal right coset if group is not abelian?
Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=1, (123)(132)$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
answered Mar 13 at 9:06
user1729user1729
17.4k64193
$endgroup$
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
add a comment |
$begingroup$
There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, circ)$, must satisfy four requirements known as the group axioms:
Closure. For all $a, bin G$ their product $a circ b$ is an element of $G$.
Associativity. For all $a, b, c in G$ we have $(a circ b) circ c = a circ (b circ c)$.
Identity element. There exists an element $e in G$ such that, for every element $a in G$, the equation $e circ a = a circ e = a$ holds.
Inverse element. For each $a in G$, there exists an element $b in G$ such that $a circ b = b circ a = e$, where $e$ is an (the) identity element.
As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=1,2,3,4,5$ but $2times3=6notin G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)
You also ask:
Can a left coset equal right coset if group is not abelian?
Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=1, (123)(132)$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
answered Mar 13 at 9:06
user1729user1729
17.4k64193
$endgroup$
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
add a comment |
$begingroup$
There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, circ)$, must satisfy four requirements known as the group axioms:
Closure. For all $a, bin G$ their product $a circ b$ is an element of $G$.
Associativity. For all $a, b, c in G$ we have $(a circ b) circ c = a circ (b circ c)$.
Identity element. There exists an element $e in G$ such that, for every element $a in G$, the equation $e circ a = a circ e = a$ holds.
Inverse element. For each $a in G$, there exists an element $b in G$ such that $a circ b = b circ a = e$, where $e$ is an (the) identity element.
As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=1,2,3,4,5$ but $2times3=6notin G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)
You also ask:
Can a left coset equal right coset if group is not abelian?
Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=1, (123)(132)$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
answered Mar 13 at 9:06
user1729user1729
17.4k64193
$endgroup$
There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to this question.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):
A group is a set, $G$, together with an operation $circ$ which may be applied to pairs of elements in $G$. To qualify as a group, the set and operation, $(G, circ)$, must satisfy four requirements known as the group axioms:
Closure. For all $a, bin G$ their product $a circ b$ is an element of $G$.
Associativity. For all $a, b, c in G$ we have $(a circ b) circ c = a circ (b circ c)$.
Identity element. There exists an element $e in G$ such that, for every element $a in G$, the equation $e circ a = a circ e = a$ holds.
Inverse element. For each $a in G$, there exists an element $b in G$ such that $a circ b = b circ a = e$, where $e$ is an (the) identity element.
As you point out, $G$ contains two (equivalence classes) of integers such that their product is not in $G$. For example, if $m=6$ then $G=1,2,3,4,5$ but $2times3=6notin G$. Hence, $G$ is not a group as it fails the "closure" axiom.
(In conclusion: your proof is fine.)
You also ask:
Can a left coset equal right coset if group is not abelian?
Yes. A subgroup $H$ is normal precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, $S_3$ is non-abelian, and if $H=1, (123)(132)$ then the left cosets are $H$ and $(12)H$, while the right cosets are $H$ and $H(12)$.
In the above example, $H$ is abelian but $G=S_3$ is not. There are other examples where both $H$ and $G$ are non-abelian. For example, if you have come across the alternating group $A_n$, then $A_n$ is a normal subgroup of $S_n$ (so left cosets=right cosets), and $A_n$ is non-abelian when $n>3$. (In the above example we actually have $H=A_3$, and this is cyclic so abelian.)
answered Mar 13 at 9:06
user1729user1729
17.4k64193
edited Mar 13 at 9:13
answered Mar 13 at 9:06
user1729user1729
17.4k64193
answered Mar 13 at 9:06
user1729user1729
17.4k64193
answered Mar 13 at 9:06
user1729user1729
17.4k64193
17.4k64193
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
add a comment |
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
$begingroup$
thanks a lot :)
$endgroup$
– Eden Hazard
Mar 13 at 18:35
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%2f3145241%2fmodulo-groups-and-non-prime-numbers%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
$begingroup$
You are on the right track, but I don't understand your last sentence. If it is a group, every element is invertible. Consider $r^-1$. What can you say now?
$endgroup$
– AnalysisStudent0414
Mar 12 at 15:47
1
$begingroup$
One may think like this that, not all elements have multiplicative inverse!
$endgroup$
– Qurultay
Mar 12 at 15:47
1
$begingroup$
You are on the right track. Let $M=acdot b$, where $a,b>1$. Then, if the multiplication operator on the modulo group is a group, then $a^-1$ exists. But then, $acdot bcdot a^-1 = 0 = acdot a^-1cdot b = b$. This contradicts $M>b>1$.
$endgroup$
– Don Thousand
Mar 12 at 16:17
$begingroup$
@DonThousand ah yes I get you thanks , but I still don't get why showing its not closed doesn't work, M is non prime so r,s must belong to the set G right?
$endgroup$
– Eden Hazard
Mar 12 at 18:36
1
$begingroup$
@EdenHazard Yes, "not closed" works. (The issue I think is that there are different, equivalent, definitions of a group. One has 4 axioms, including "closure", while the second has 3 axioms and disguises closure with the words "binary relation".)
$endgroup$
– user1729
Mar 13 at 8:48