Yhyjyk

What are some good books on Machine Learning and AI like Krugman, Wells and Graddy's "Essentials of Economics"

Is it inappropriate for a student to attend their mentor's dissertation defense?

Should I tell management that I intend to leave due to bad software development practices?

What is the idiomatic way to say "clothing fits"?

Ambiguity in the definition of entropy

Arrow those variables!

Is this a hacking script in function.php?

Personal Teleportation: From Rags to Riches

Why is this clock signal connected to a capacitor to gnd?

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

One verb to replace 'be a member of' a club

Could the museum Saturn V's be refitted for one more flight?

Method Does Not Exist error message

Alternative to sending password over mail?

Why does this cyclic subgroup have only 4 subgroups?

Forming a German sentence with/without the verb at the end

What type of content (depth/breadth) is expected for a short presentation for Asst Professor interview in the UK?

Why would the Red Woman birth a shadow if she worshipped the Lord of the Light?

Can the Meissner effect explain very large floating structures?

GFCI outlets - can they be repaired? Are they really needed at the end of a circuit?

What reasons are there for a Capitalist to oppose a 100% inheritance tax?

What does the expression "A Mann!" means

iPad being using in wall mount battery swollen

How do I deal with an unproductive colleague in a small company?

Why does this cyclic subgroup have only 4 subgroups?

What does it mean to have no proper non-trivial subgroupCyclic subgroup of a cyclic groupProof on Cyclic Subgroup GenerationIf $G$ has only 2 non-trivial proper subgroups H, N, then H, N are cyclic subgroup of $G$.Number of cyclic subgroups of the alternating group $A_8$All groups of order 10 have a proper normal subgroupHow many subgroups of order 17 does $S_17$ have?Why do Sylow $3$-subgroups intersect only in the identity?Group with proper subgroups infinite cyclicHow many noncyclic submodules with $9$ elements does $V$ have?

1

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
  $endgroup$
  – Minus One-Twelfth
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why? Could you help me understand how you got to that conclusion?
  $endgroup$
  – Evan Kim
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $1,a^5$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^10=a^4$
  $endgroup$
  – J. W. Tanner
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
  $endgroup$
  – Minus One-Twelfth
  55 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
  $endgroup$
  – Minus One-Twelfth
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why? Could you help me understand how you got to that conclusion?
  $endgroup$
  – Evan Kim
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $1,a^5$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^10=a^4$
  $endgroup$
  – J. W. Tanner
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
  $endgroup$
  – Minus One-Twelfth
  55 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

$endgroup$

share|cite|improve this question

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

share|cite|improve this question

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

edited 2 hours ago

J. W. Tanner

4,3651320

edited 2 hours ago

J. W. Tanner

4,3651320

asked 2 hours ago

Evan KimEvan Kim

66319

asked 2 hours ago

Evan KimEvan Kim

66319

4 Answers
4

active

oldest

votes

2

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered 2 hours ago

PeterPeter

48.9k1240137

$endgroup$

add a comment | 

2

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

$endgroup$

add a comment | 

1

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

$endgroup$

add a comment | 

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered 2 hours ago

ShaunShaun

10.1k113685

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3173761%2fwhy-does-this-cyclic-subgroup-have-only-4-subgroups%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered 2 hours ago

PeterPeter

48.9k1240137

$endgroup$

add a comment | 

2

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered 2 hours ago

PeterPeter

48.9k1240137

$endgroup$

add a comment | 

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered 2 hours ago

PeterPeter

48.9k1240137

$endgroup$

share|cite|improve this answer

answered 2 hours ago

PeterPeter

48.9k1240137

answered 2 hours ago

PeterPeter

48.9k1240137

answered 2 hours ago

PeterPeter

48.9k1240137

2

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

$endgroup$

add a comment | 

2

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

$endgroup$

add a comment | 

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

$endgroup$

share|cite|improve this answer

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

answered 2 hours ago

TheSilverDoeTheSilverDoe

5,157215

1

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

$endgroup$

add a comment | 

1

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

$endgroup$

add a comment | 

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

answered 1 hour ago

Jack PfaffingerJack Pfaffinger

3841112

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered 2 hours ago

ShaunShaun

10.1k113685

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered 2 hours ago

ShaunShaun

10.1k113685

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered 2 hours ago

ShaunShaun

10.1k113685

$endgroup$

share|cite|improve this answer

answered 2 hours ago

ShaunShaun

10.1k113685

answered 2 hours ago

ShaunShaun

10.1k113685

answered 2 hours ago

ShaunShaun

10.1k113685