Mixing
Example of $a,~bin G$ such that $abin Hleq G$ and $a^2b^2notin H.$
$begingroup$
Let $G$ be a group and $H$ be a subgroup of $G$. Let also $a,~bin G$ such that $abin H$.
True or false? $a^2b^2in H.$
Attempt. I believe the answer is no (i have proved that the statement is true for normal subgroups, but it seems that there is no need to hold for arbitrary subgroups). I was looking for a counterexample in a non abelian group of small order, such as $S_3$, or $S_4$, but i couldn't find a suitable combination of $Hleq S_n$, $sigma$ and $tauin S_n$ such that $sigma tau in H$ and $sigma^2 tau^2 notin H.$
Thanks in advance for the help.
abstract-algebra group-theory permutations examples-counterexamples
edited Jan 11 at 13:51
Shaun
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group and $H$ be a subgroup of $G$. Let also $a,~bin G$ such that $abin H$.
True or false? $a^2b^2in H.$
Attempt. I believe the answer is no (i have proved that the statement is true for normal subgroups, but it seems that there is no need to hold for arbitrary subgroups). I was looking for a counterexample in a non abelian group of small order, such as $S_3$, or $S_4$, but i couldn't find a suitable combination of $Hleq S_n$, $sigma$ and $tauin S_n$ such that $sigma tau in H$ and $sigma^2 tau^2 notin H.$
Thanks in advance for the help.
abstract-algebra group-theory permutations examples-counterexamples
edited Jan 11 at 13:51
Shaun
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
$endgroup$
1
$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15
add a comment |
$begingroup$
Let $G$ be a group and $H$ be a subgroup of $G$. Let also $a,~bin G$ such that $abin H$.
True or false? $a^2b^2in H.$
Attempt. I believe the answer is no (i have proved that the statement is true for normal subgroups, but it seems that there is no need to hold for arbitrary subgroups). I was looking for a counterexample in a non abelian group of small order, such as $S_3$, or $S_4$, but i couldn't find a suitable combination of $Hleq S_n$, $sigma$ and $tauin S_n$ such that $sigma tau in H$ and $sigma^2 tau^2 notin H.$
Thanks in advance for the help.
abstract-algebra group-theory permutations examples-counterexamples
edited Jan 11 at 13:51
Shaun
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
$endgroup$
Let $G$ be a group and $H$ be a subgroup of $G$. Let also $a,~bin G$ such that $abin H$.
True or false? $a^2b^2in H.$
Attempt. I believe the answer is no (i have proved that the statement is true for normal subgroups, but it seems that there is no need to hold for arbitrary subgroups). I was looking for a counterexample in a non abelian group of small order, such as $S_3$, or $S_4$, but i couldn't find a suitable combination of $Hleq S_n$, $sigma$ and $tauin S_n$ such that $sigma tau in H$ and $sigma^2 tau^2 notin H.$
Thanks in advance for the help.
abstract-algebra group-theory permutations examples-counterexamples
abstract-algebra group-theory permutations examples-counterexamples
edited Jan 11 at 13:51
Shaun
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
edited Jan 11 at 13:51
Shaun
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
edited Jan 11 at 13:51
Shaun
9,065113683
edited Jan 11 at 13:51
Shaun
9,065113683
edited Jan 11 at 13:51
Shaun
9,065113683
9,065113683
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
asked Jan 9 at 22:39
Nikolaos SkoutNikolaos Skout
2,249618
2,249618
1
$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15
add a comment |
1
$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15
1
1
$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15
$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider $S_3$.
Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$
Let $H={1, ab}$. Then $abin H$ but $a^2b^2notin H$
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
$endgroup$
add a comment |
$begingroup$
Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.
Take $H$ to be the cyclic subgroup generated by $ab$.
The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n ge 1$ and $(b^{-1}a^{-1})^n$ for $n ge 1$.
Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 notin H$.
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
$endgroup$
add a comment |
$begingroup$
Let $u in G$, $v in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab in H$.
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 in H Leftrightarrow uvu^{-1}v in H Leftrightarrow uvu^{-1} in H$.
Thus if $H$ is not normal, the property does not hold.
answered Jan 9 at 23:01
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $S_3$.
Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$
Let $H={1, ab}$. Then $abin H$ but $a^2b^2notin H$
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
$endgroup$
add a comment |
$begingroup$
Consider $S_3$.
Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$
Let $H={1, ab}$. Then $abin H$ but $a^2b^2notin H$
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
$endgroup$
add a comment |
$begingroup$
Consider $S_3$.
Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$
Let $H={1, ab}$. Then $abin H$ but $a^2b^2notin H$
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
$endgroup$
Consider $S_3$.
Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$
Let $H={1, ab}$. Then $abin H$ but $a^2b^2notin H$
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
answered Jan 9 at 22:54
John DoumaJohn Douma
5,51211319
5,51211319
add a comment |
add a comment |
$begingroup$
Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.
Take $H$ to be the cyclic subgroup generated by $ab$.
The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n ge 1$ and $(b^{-1}a^{-1})^n$ for $n ge 1$.
Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 notin H$.
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
$endgroup$
add a comment |
$begingroup$
Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.
Take $H$ to be the cyclic subgroup generated by $ab$.
The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n ge 1$ and $(b^{-1}a^{-1})^n$ for $n ge 1$.
Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 notin H$.
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
$endgroup$
add a comment |
$begingroup$
Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.
Take $H$ to be the cyclic subgroup generated by $ab$.
The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n ge 1$ and $(b^{-1}a^{-1})^n$ for $n ge 1$.
Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 notin H$.
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
$endgroup$
Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.
Take $H$ to be the cyclic subgroup generated by $ab$.
The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n ge 1$ and $(b^{-1}a^{-1})^n$ for $n ge 1$.
Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 notin H$.
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
answered Jan 9 at 22:47
Lee MosherLee Mosher
49.1k33685
49.1k33685
add a comment |
add a comment |
$begingroup$
Let $u in G$, $v in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab in H$.
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 in H Leftrightarrow uvu^{-1}v in H Leftrightarrow uvu^{-1} in H$.
Thus if $H$ is not normal, the property does not hold.
answered Jan 9 at 23:01
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
add a comment |
$begingroup$
Let $u in G$, $v in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab in H$.
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 in H Leftrightarrow uvu^{-1}v in H Leftrightarrow uvu^{-1} in H$.
Thus if $H$ is not normal, the property does not hold.
answered Jan 9 at 23:01
MindlackMindlack
3,53717
$endgroup$
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
add a comment |
$begingroup$
Let $u in G$, $v in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab in H$.
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 in H Leftrightarrow uvu^{-1}v in H Leftrightarrow uvu^{-1} in H$.
Thus if $H$ is not normal, the property does not hold.
answered Jan 9 at 23:01
MindlackMindlack
3,53717
$endgroup$
Let $u in G$, $v in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab in H$.
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 in H Leftrightarrow uvu^{-1}v in H Leftrightarrow uvu^{-1} in H$.
Thus if $H$ is not normal, the property does not hold.
answered Jan 9 at 23:01
MindlackMindlack
3,53717
answered Jan 9 at 23:01
MindlackMindlack
3,53717
answered Jan 9 at 23:01
MindlackMindlack
3,53717
answered Jan 9 at 23:01
MindlackMindlack
3,53717
3,53717
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
add a comment |
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Failure of normality does not imply that for all $u in G$, $nu in H$ we have $unu u^{-1} notin H$.
$endgroup$
– Lee Mosher
Jan 9 at 23:04
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ».
$endgroup$
– Mindlack
Jan 9 at 23:09
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
$begingroup$
+1, to me this is just best possible.
$endgroup$
– Andreas Caranti
Jan 10 at 11:23
add a comment |
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$begingroup$
Just out of curiosity, what pair $sigma,tauin S_3$ did you try in your ques for an example with $sigma^2tau^2notinlanglesigmataurangle$? (I assume they did not commute, and were not both of order $2$.)
$endgroup$
– bof
Jan 9 at 23:15