Why does this cyclic subgroup have only 4 subgroups?What does it mean to have no proper non-trivial...
Should I cover my bicycle overnight while bikepacking?
Arrow those variables!
What exploit Are these user agents trying to use?
Why would the Red Woman birth a shadow if she worshipped the Lord of the Light?
Why didn't Boeing produce its own regional jet?
How dangerous is XSS?
Venezuelan girlfriend wants to travel the USA to be with me. What is the process?
Why is this clock signal connected to a capacitor to gnd?
If human space travel is limited by the G force vulnerability, is there a way to counter G forces?
What method can I use to design a dungeon difficult enough that the PCs can't make it through without killing them?
What type of content (depth/breadth) is expected for a short presentation for Asst Professor interview in the UK?
Why no variance term in Bayesian logistic regression?
Could the museum Saturn V's be refitted for one more flight?
How badly should I try to prevent a user from XSSing themselves?
Is it possible to create a QR code using text?
Is it inappropriate for a student to attend their mentor's dissertation defense?
Mathematica command that allows it to read my intentions
Am I breaking OOP practice with this architecture?
How can saying a song's name be a copyright violation?
Why do bosons tend to occupy the same state?
Are there any examples of a variable being normally distributed that is *not* due to the Central Limit Theorem?
Which is the best way to check return result?
How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?
How much of data wrangling is a data scientist's job?
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?
$begingroup$
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
edited 2 hours ago
J. W. Tanner
4,3651320
asked 2 hours ago
Evan KimEvan Kim
66319
$endgroup$
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
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
1 hour 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
edited 2 hours ago
J. W. Tanner
4,3651320
asked 2 hours ago
Evan KimEvan Kim
66319
$endgroup$
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
abstract-algebra group-theory
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
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
edited 2 hours ago
J. W. Tanner
4,3651320
4,3651320
asked 2 hours ago
Evan KimEvan Kim
66319
asked 2 hours ago
Evan KimEvan Kim
66319
asked 2 hours ago
Evan KimEvan Kim
66319
66319
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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
1 hour ago
$begingroup$
The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
$endgroup$
– Minus One-Twelfth
2 hours ago
$begingroup$
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
$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
2 hours ago
2
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
$begingroup$
${1,a^5}$ 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
1 hour 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
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$[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.
answered 2 hours ago
PeterPeter
48.9k1240137
$endgroup$
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.
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
$endgroup$
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 $a$.
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
$endgroup$
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 $H$.
answered 2 hours ago
ShaunShaun
10.1k113685
$endgroup$
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours 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
});
}
});
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%2f3173761%2fwhy-does-this-cyclic-subgroup-have-only-4-subgroups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$[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.
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.
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.
answered 2 hours ago
PeterPeter
48.9k1240137
$endgroup$
$[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.
answered 2 hours ago
PeterPeter
48.9k1240137
answered 2 hours ago
PeterPeter
48.9k1240137
answered 2 hours ago
PeterPeter
48.9k1240137
answered 2 hours ago
PeterPeter
48.9k1240137
48.9k1240137
add a comment |
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.
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.
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.
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
$endgroup$
$lbrace 1, a^5 rbrace$ 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.
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
answered 2 hours ago
TheSilverDoeTheSilverDoe
5,157215
5,157215
add a comment |
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 $a$.
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 $a$.
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 $a$.
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
$endgroup$
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 $a$.
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
answered 1 hour ago
Jack PfaffingerJack Pfaffinger
3841112
3841112
add a comment |
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 $H$.
answered 2 hours ago
ShaunShaun
10.1k113685
$endgroup$
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours 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 $H$.
answered 2 hours ago
ShaunShaun
10.1k113685
$endgroup$
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours 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 $H$.
answered 2 hours ago
ShaunShaun
10.1k113685
$endgroup$
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 $H$.
answered 2 hours ago
ShaunShaun
10.1k113685
answered 2 hours ago
ShaunShaun
10.1k113685
answered 2 hours ago
ShaunShaun
10.1k113685
answered 2 hours ago
ShaunShaun
10.1k113685
10.1k113685
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours ago
add a comment |
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours ago
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours ago
$begingroup$
Why the downvote?
$endgroup$
– Shaun
2 hours ago
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%2f3173761%2fwhy-does-this-cyclic-subgroup-have-only-4-subgroups%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
$begingroup$
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
1 hour ago