Analyzing group form given generator relation
up vote
0
down vote
favorite
Given $G$ be any finite group and $a,bin G$ s.t $o(a)=7$ and $O(b)=49$ with the condition $aba^{-1}=b^8$ then consider the following statement :
a)$G$ must be an abelian group
b)If $a,b$ generate the group $G$ then order of $G$ is $7^3$
c)$7^3$ is the order of the proper subgroup of $G$
.-------------
Option a is not true by using contradiction but what about others option ???
group-theory
edited Nov 27 at 16:58
Shubham
1,5941519
asked Nov 27 at 12:30
G C R
265
add a comment |
up vote
0
down vote
favorite
Given $G$ be any finite group and $a,bin G$ s.t $o(a)=7$ and $O(b)=49$ with the condition $aba^{-1}=b^8$ then consider the following statement :
a)$G$ must be an abelian group
b)If $a,b$ generate the group $G$ then order of $G$ is $7^3$
c)$7^3$ is the order of the proper subgroup of $G$
.-------------
Option a is not true by using contradiction but what about others option ???
group-theory
edited Nov 27 at 16:58
Shubham
1,5941519
asked Nov 27 at 12:30
G C R
265
1
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $G$ be any finite group and $a,bin G$ s.t $o(a)=7$ and $O(b)=49$ with the condition $aba^{-1}=b^8$ then consider the following statement :
a)$G$ must be an abelian group
b)If $a,b$ generate the group $G$ then order of $G$ is $7^3$
c)$7^3$ is the order of the proper subgroup of $G$
.-------------
Option a is not true by using contradiction but what about others option ???
group-theory
edited Nov 27 at 16:58
Shubham
1,5941519
asked Nov 27 at 12:30
G C R
265
Given $G$ be any finite group and $a,bin G$ s.t $o(a)=7$ and $O(b)=49$ with the condition $aba^{-1}=b^8$ then consider the following statement :
a)$G$ must be an abelian group
b)If $a,b$ generate the group $G$ then order of $G$ is $7^3$
c)$7^3$ is the order of the proper subgroup of $G$
.-------------
Option a is not true by using contradiction but what about others option ???
group-theory
group-theory
edited Nov 27 at 16:58
Shubham
1,5941519
asked Nov 27 at 12:30
G C R
265
edited Nov 27 at 16:58
Shubham
1,5941519
asked Nov 27 at 12:30
G C R
265
edited Nov 27 at 16:58
Shubham
1,5941519
edited Nov 27 at 16:58
Shubham
1,5941519
edited Nov 27 at 16:58
Shubham
1,5941519
1,5941519
asked Nov 27 at 12:30
G C R
265
asked Nov 27 at 12:30
G C R
265
asked Nov 27 at 12:30
G C R
265
265
1
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43
add a comment |
1
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43
1
1
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
So $G= langle a,b| a^7,b^{49}, aba^{-1}b^{-8}rangle$.
The third relator says that any element in $G$ can be written in the form $a^kb^m$. So $|G|leq 7^{3}$. Further more $|G|>49$ as the order of $b$ is $49$, and therefore 49 divides $|G|$. So you have $7^3$ possible choices for the elements in $G$.
answered Nov 27 at 12:38
mathnoob
1,646322
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
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',
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%2f3015711%2fanalyzing-group-form-given-generator-relation%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
up vote
2
down vote
accepted
So $G= langle a,b| a^7,b^{49}, aba^{-1}b^{-8}rangle$.
The third relator says that any element in $G$ can be written in the form $a^kb^m$. So $|G|leq 7^{3}$. Further more $|G|>49$ as the order of $b$ is $49$, and therefore 49 divides $|G|$. So you have $7^3$ possible choices for the elements in $G$.
answered Nov 27 at 12:38
mathnoob
1,646322
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
add a comment |
up vote
2
down vote
accepted
So $G= langle a,b| a^7,b^{49}, aba^{-1}b^{-8}rangle$.
The third relator says that any element in $G$ can be written in the form $a^kb^m$. So $|G|leq 7^{3}$. Further more $|G|>49$ as the order of $b$ is $49$, and therefore 49 divides $|G|$. So you have $7^3$ possible choices for the elements in $G$.
answered Nov 27 at 12:38
mathnoob
1,646322
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
So $G= langle a,b| a^7,b^{49}, aba^{-1}b^{-8}rangle$.
The third relator says that any element in $G$ can be written in the form $a^kb^m$. So $|G|leq 7^{3}$. Further more $|G|>49$ as the order of $b$ is $49$, and therefore 49 divides $|G|$. So you have $7^3$ possible choices for the elements in $G$.
answered Nov 27 at 12:38
mathnoob
1,646322
So $G= langle a,b| a^7,b^{49}, aba^{-1}b^{-8}rangle$.
The third relator says that any element in $G$ can be written in the form $a^kb^m$. So $|G|leq 7^{3}$. Further more $|G|>49$ as the order of $b$ is $49$, and therefore 49 divides $|G|$. So you have $7^3$ possible choices for the elements in $G$.
answered Nov 27 at 12:38
mathnoob
1,646322
edited Nov 27 at 12:55
answered Nov 27 at 12:38
mathnoob
1,646322
answered Nov 27 at 12:38
mathnoob
1,646322
answered Nov 27 at 12:38
mathnoob
1,646322
1,646322
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
add a comment |
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
1
1
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
The first relator should be $a^7$, not $a^8$. Your argument proves that $|G| le 7^3$ rather than $|G|=7^3$, although the assumptions about orders of $a$ and $b$ in the question suggests that we are allowed to assume that $|G|>49$, and hence $|G| = 7^3$.
– Derek Holt
Nov 27 at 12:48
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Oh right Thank you! I edited it.
– mathnoob
Nov 27 at 12:51
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
Sorry, I'm actually confused now, why can't the order of the group be $49*3$ for example? I know that The group could have at least $2$ left $langle b rangle$ cosets, namely $a<b>, <b>$. which gives $2*49$ elements. Why more?
– mathnoob
Nov 27 at 22:29
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
We are told that $o(a)=7$ and $o(b)=49$. So $b ne b^8$, and hence $a$ does not centralize $b$. In fact no nontirival power of $a$ can centralize $b$, and so $langle a rangle cap langle b rangle = 1$ and hence $|langle a,b rangle| ge |langle a rangle||langle b rangle| = 7^3$.
– Derek Holt
Nov 27 at 23:02
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%2f3015711%2fanalyzing-group-form-given-generator-relation%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
Some helpful suggestions: (1) Spelling matters. I think you meant "statements", rather than "statement" but what you wrote literally suggests we take (a), (b), and (c) together as a compound statement. (2) Don't use shorthand unless you know it is correct shorthand. Numbers aren't groups, but $7^3$ can't be used as shorthand for any group of order $7^3$ because by convention it refers specifically to the elementary abelian group of that order (the direct product three cyclic groups of order $7$). If that's not what you meant by it, then you asked a different question that you meant to ask.
– C Monsour
Nov 27 at 12:43