Mixing
![Delayed job joining [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjbpfN9tAutmK93bJRC3ZoROZzi2TJDms5n8_qJuhgE0a9b52OOHayv3NGT8igAdFL7byXNst-_1DZK5SjrIJ28_6RQPUpBROqMs5s6jo-ZsjX8kjDwfxJufIitH3TaQRXWaGSQKRQib-f/s72-c/1.jpg)
Intersection of two subgroups
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $G$ be a group and $H$, $K$ two cyclic subgroups of $G$ of the same order such that:$$
H=langle xrangle,~~K=langle yrangle~~ textand~~xneq y.$$
Can we said that $Hcap K$ is trivial?
Thank you
abstract-algebra group-theory cyclic-groups
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
asked Aug 7 at 17:40
Amine El Bouzidi
344
add a comment |Â
up vote
2
down vote
favorite
Let $G$ be a group and $H$, $K$ two cyclic subgroups of $G$ of the same order such that:$$
H=langle xrangle,~~K=langle yrangle~~ textand~~xneq y.$$
Can we said that $Hcap K$ is trivial?
Thank you
abstract-algebra group-theory cyclic-groups
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
asked Aug 7 at 17:40
Amine El Bouzidi
344
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G$ be a group and $H$, $K$ two cyclic subgroups of $G$ of the same order such that:$$
H=langle xrangle,~~K=langle yrangle~~ textand~~xneq y.$$
Can we said that $Hcap K$ is trivial?
Thank you
abstract-algebra group-theory cyclic-groups
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
asked Aug 7 at 17:40
Amine El Bouzidi
344
Let $G$ be a group and $H$, $K$ two cyclic subgroups of $G$ of the same order such that:$$
H=langle xrangle,~~K=langle yrangle~~ textand~~xneq y.$$
Can we said that $Hcap K$ is trivial?
Thank you
abstract-algebra group-theory cyclic-groups
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
asked Aug 7 at 17:40
Amine El Bouzidi
344
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
edited Aug 7 at 17:47
Arnaud Mortier
19.6k22159
19.6k22159
asked Aug 7 at 17:40
Amine El Bouzidi
344
asked Aug 7 at 17:40
Amine El Bouzidi
344
asked Aug 7 at 17:40
Amine El Bouzidi
344
344
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58
add a comment |Â
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
No we can't: consider $langle 1rangle=langle 2rangle$ in $Bbb Z/3Bbb Z$.
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
add a comment |Â
up vote
1
down vote
No, let $H=langle 2 rangle$ and $K=langle 3 rangle$ as subgroups of $mathbbZ$. Then $H cap K = langle 6 rangle $.
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
add a comment |Â
up vote
1
down vote
Another example:
$langle 1 rangle=langle -1 rangle$ in $BbbZ$. So clearly $1neq -1$ but $langle 1 ranglecaplangle -1 rangleneq e$.
By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.
answered Aug 8 at 0:50
Alan Wang
4,559932
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
No we can't: consider $langle 1rangle=langle 2rangle$ in $Bbb Z/3Bbb Z$.
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
add a comment |Â
up vote
5
down vote
No we can't: consider $langle 1rangle=langle 2rangle$ in $Bbb Z/3Bbb Z$.
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
add a comment |Â
up vote
5
down vote
up vote
5
down vote
No we can't: consider $langle 1rangle=langle 2rangle$ in $Bbb Z/3Bbb Z$.
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
No we can't: consider $langle 1rangle=langle 2rangle$ in $Bbb Z/3Bbb Z$.
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
answered Aug 7 at 17:46
Arnaud Mortier
19.6k22159
19.6k22159
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
add a comment |Â
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
Thank you Arnaud for your answer.
– Amine El Bouzidi
Aug 9 at 12:24
add a comment |Â
up vote
1
down vote
No, let $H=langle 2 rangle$ and $K=langle 3 rangle$ as subgroups of $mathbbZ$. Then $H cap K = langle 6 rangle $.
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
add a comment |Â
up vote
1
down vote
No, let $H=langle 2 rangle$ and $K=langle 3 rangle$ as subgroups of $mathbbZ$. Then $H cap K = langle 6 rangle $.
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, let $H=langle 2 rangle$ and $K=langle 3 rangle$ as subgroups of $mathbbZ$. Then $H cap K = langle 6 rangle $.
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
No, let $H=langle 2 rangle$ and $K=langle 3 rangle$ as subgroups of $mathbbZ$. Then $H cap K = langle 6 rangle $.
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
answered Aug 7 at 18:29
Nicky Hekster
27.1k53152
27.1k53152
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
add a comment |Â
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
Thank you Nicky for your answer
– Amine El Bouzidi
Aug 9 at 12:24
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
No problem Amine, hope you learned something from all the answers.
– Nicky Hekster
Aug 9 at 12:59
add a comment |Â
up vote
1
down vote
Another example:
$langle 1 rangle=langle -1 rangle$ in $BbbZ$. So clearly $1neq -1$ but $langle 1 ranglecaplangle -1 rangleneq e$.
By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.
answered Aug 8 at 0:50
Alan Wang
4,559932
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
add a comment |Â
up vote
1
down vote
Another example:
$langle 1 rangle=langle -1 rangle$ in $BbbZ$. So clearly $1neq -1$ but $langle 1 ranglecaplangle -1 rangleneq e$.
By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.
answered Aug 8 at 0:50
Alan Wang
4,559932
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another example:
$langle 1 rangle=langle -1 rangle$ in $BbbZ$. So clearly $1neq -1$ but $langle 1 ranglecaplangle -1 rangleneq e$.
By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.
answered Aug 8 at 0:50
Alan Wang
4,559932
Another example:
$langle 1 rangle=langle -1 rangle$ in $BbbZ$. So clearly $1neq -1$ but $langle 1 ranglecaplangle -1 rangleneq e$.
By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.
answered Aug 8 at 0:50
Alan Wang
4,559932
answered Aug 8 at 0:50
Alan Wang
4,559932
answered Aug 8 at 0:50
Alan Wang
4,559932
answered Aug 8 at 0:50
Alan Wang
4,559932
4,559932
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
add a comment |Â
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
thank you Alan for your answer
– Amine El Bouzidi
Aug 9 at 12:23
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2875191%2fintersection-of-two-subgroups%23new-answer', 'question_page');
);
Post as a guest
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
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
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
In fact, it's false.
– Rafael Gonzalez Lopez
Aug 7 at 17:44
You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $mathbb Z_p^2timesmathbb Z_p$ consider $langle (1,0)rangle$ versus $langle(1,1)rangle$.
– Henning Makholm
Aug 7 at 17:54
Of course you can say it if you want to, but it would be false.
– Derek Holt
Aug 7 at 19:29
Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration.
– Alan Wang
Aug 8 at 13:58