If A is permutable subgroup with B , then A is permutable with B's conjugate?
up vote
-1
down vote
favorite
Let $G$ be a finite group and $A,B, C le G$ subgroups.
How to prove that if $AB = BA$ ( permutable subgroups ) , and $C$ is conjugate of $B$ Then $AC = CA$ ? If not , can you give me a counter example?
Edit: Just need to check if my solution is correct.
Well I guess the statement is true for abelian groups since we have
If G is abelian, then $gag^{-1}= a$ for all a and g in G; so Cl(a) = {a} for all a in G.
so If $A$ and $B$ are permutable and $C$ is the conjugate of $B$ ( which equals $B$ ) then $A$ and $C$ are also perumtable.
but if the group was not abelian , the statement need not to be true.
and my example is
Let $G=$ $D_{3}= langle r,s mid r^3=s^2=1 , rs= sr^2 rangle$
and Let $B$ = $A = langle s rangle$
Then $AB = BA$.
Now the conjugate of $B$ : $C = langle sr rangle$ is not permutable with A since $$ AC= {1 , s , sr , r}$$
and $$ CA= {1 , s , sr , r^2 }$$
abstract-algebra group-theory finite-groups
asked Nov 27 at 3:58
H.koby
407
add a comment |
up vote
-1
down vote
favorite
Let $G$ be a finite group and $A,B, C le G$ subgroups.
How to prove that if $AB = BA$ ( permutable subgroups ) , and $C$ is conjugate of $B$ Then $AC = CA$ ? If not , can you give me a counter example?
Edit: Just need to check if my solution is correct.
Well I guess the statement is true for abelian groups since we have
If G is abelian, then $gag^{-1}= a$ for all a and g in G; so Cl(a) = {a} for all a in G.
so If $A$ and $B$ are permutable and $C$ is the conjugate of $B$ ( which equals $B$ ) then $A$ and $C$ are also perumtable.
but if the group was not abelian , the statement need not to be true.
and my example is
Let $G=$ $D_{3}= langle r,s mid r^3=s^2=1 , rs= sr^2 rangle$
and Let $B$ = $A = langle s rangle$
Then $AB = BA$.
Now the conjugate of $B$ : $C = langle sr rangle$ is not permutable with A since $$ AC= {1 , s , sr , r}$$
and $$ CA= {1 , s , sr , r^2 }$$
abstract-algebra group-theory finite-groups
asked Nov 27 at 3:58
H.koby
407
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $G$ be a finite group and $A,B, C le G$ subgroups.
How to prove that if $AB = BA$ ( permutable subgroups ) , and $C$ is conjugate of $B$ Then $AC = CA$ ? If not , can you give me a counter example?
Edit: Just need to check if my solution is correct.
Well I guess the statement is true for abelian groups since we have
If G is abelian, then $gag^{-1}= a$ for all a and g in G; so Cl(a) = {a} for all a in G.
so If $A$ and $B$ are permutable and $C$ is the conjugate of $B$ ( which equals $B$ ) then $A$ and $C$ are also perumtable.
but if the group was not abelian , the statement need not to be true.
and my example is
Let $G=$ $D_{3}= langle r,s mid r^3=s^2=1 , rs= sr^2 rangle$
and Let $B$ = $A = langle s rangle$
Then $AB = BA$.
Now the conjugate of $B$ : $C = langle sr rangle$ is not permutable with A since $$ AC= {1 , s , sr , r}$$
and $$ CA= {1 , s , sr , r^2 }$$
abstract-algebra group-theory finite-groups
asked Nov 27 at 3:58
H.koby
407
Let $G$ be a finite group and $A,B, C le G$ subgroups.
How to prove that if $AB = BA$ ( permutable subgroups ) , and $C$ is conjugate of $B$ Then $AC = CA$ ? If not , can you give me a counter example?
Edit: Just need to check if my solution is correct.
Well I guess the statement is true for abelian groups since we have
If G is abelian, then $gag^{-1}= a$ for all a and g in G; so Cl(a) = {a} for all a in G.
so If $A$ and $B$ are permutable and $C$ is the conjugate of $B$ ( which equals $B$ ) then $A$ and $C$ are also perumtable.
but if the group was not abelian , the statement need not to be true.
and my example is
Let $G=$ $D_{3}= langle r,s mid r^3=s^2=1 , rs= sr^2 rangle$
and Let $B$ = $A = langle s rangle$
Then $AB = BA$.
Now the conjugate of $B$ : $C = langle sr rangle$ is not permutable with A since $$ AC= {1 , s , sr , r}$$
and $$ CA= {1 , s , sr , r^2 }$$
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Nov 27 at 3:58
H.koby
407
asked Nov 27 at 3:58
H.koby
407
edited Nov 27 at 18:21
asked Nov 27 at 3:58
H.koby
407
asked Nov 27 at 3:58
H.koby
407
asked Nov 27 at 3:58
H.koby
407
407
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37
add a comment |
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3015315%2fif-a-is-permutable-subgroup-with-b-then-a-is-permutable-with-bs-conjugate%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
That sounds rather unlikely to me.
– Lord Shark the Unknown
Nov 27 at 4:04
Can you give me an example please?
– H.koby
Nov 27 at 4:05
Hint: look for an example with $A=B$.
– Derek Holt
Nov 27 at 7:37