Mixing
Example for subgroups $H$ and $K$ where $HK = K H$ and neither $H$ nor $K$ is normal?
Let $G$ be a finite group with proper non-trivial subgroups $H$ and $K$.
If $H K = K H$, then $H K$ is obviously a subgroup of $G$. It is well-known that $H$ or $K$ being normal implies that $H K$ is such a subgroup.
Question: Is there a finite group $G$ with proper non-trivial subgroups $H$ and $K$ such that $H K = K H$ and neither $H$ nor $K$ is normal in $G$?
group-theory finite-groups examples-counterexamples normal-subgroups
asked Jan 19 '18 at 18:36
Moritz
1,2031622
add a comment |
Let $G$ be a finite group with proper non-trivial subgroups $H$ and $K$.
If $H K = K H$, then $H K$ is obviously a subgroup of $G$. It is well-known that $H$ or $K$ being normal implies that $H K$ is such a subgroup.
Question: Is there a finite group $G$ with proper non-trivial subgroups $H$ and $K$ such that $H K = K H$ and neither $H$ nor $K$ is normal in $G$?
group-theory finite-groups examples-counterexamples normal-subgroups
asked Jan 19 '18 at 18:36
Moritz
1,2031622
6
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
3
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45
add a comment |
Let $G$ be a finite group with proper non-trivial subgroups $H$ and $K$.
If $H K = K H$, then $H K$ is obviously a subgroup of $G$. It is well-known that $H$ or $K$ being normal implies that $H K$ is such a subgroup.
Question: Is there a finite group $G$ with proper non-trivial subgroups $H$ and $K$ such that $H K = K H$ and neither $H$ nor $K$ is normal in $G$?
group-theory finite-groups examples-counterexamples normal-subgroups
asked Jan 19 '18 at 18:36
Moritz
1,2031622
Let $G$ be a finite group with proper non-trivial subgroups $H$ and $K$.
If $H K = K H$, then $H K$ is obviously a subgroup of $G$. It is well-known that $H$ or $K$ being normal implies that $H K$ is such a subgroup.
Question: Is there a finite group $G$ with proper non-trivial subgroups $H$ and $K$ such that $H K = K H$ and neither $H$ nor $K$ is normal in $G$?
group-theory finite-groups examples-counterexamples normal-subgroups
group-theory finite-groups examples-counterexamples normal-subgroups
asked Jan 19 '18 at 18:36
Moritz
1,2031622
asked Jan 19 '18 at 18:36
Moritz
1,2031622
asked Jan 19 '18 at 18:36
Moritz
1,2031622
asked Jan 19 '18 at 18:36
Moritz
1,2031622
asked Jan 19 '18 at 18:36
Moritz
1,2031622
1,2031622
6
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
3
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45
add a comment |
6
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
3
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45
6
6
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
3
3
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45
add a comment |
1 Answer
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Example In $D_4$ $H=<s>$,$K=<r^2s> $ Clearly HK is subgroup of G But None of H and K is normal subgroup.
answered Nov 21 '18 at 4:02
Shubham
1,5951519
add a comment |
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Example In $D_4$ $H=<s>$,$K=<r^2s> $ Clearly HK is subgroup of G But None of H and K is normal subgroup.
answered Nov 21 '18 at 4:02
Shubham
1,5951519
add a comment |
Example In $D_4$ $H=<s>$,$K=<r^2s> $ Clearly HK is subgroup of G But None of H and K is normal subgroup.
answered Nov 21 '18 at 4:02
Shubham
1,5951519
add a comment |
Example In $D_4$ $H=<s>$,$K=<r^2s> $ Clearly HK is subgroup of G But None of H and K is normal subgroup.
answered Nov 21 '18 at 4:02
Shubham
1,5951519
Example In $D_4$ $H=<s>$,$K=<r^2s> $ Clearly HK is subgroup of G But None of H and K is normal subgroup.
answered Nov 21 '18 at 4:02
Shubham
1,5951519
answered Nov 21 '18 at 4:02
Shubham
1,5951519
answered Nov 21 '18 at 4:02
Shubham
1,5951519
answered Nov 21 '18 at 4:02
Shubham
1,5951519
1,5951519
add a comment |
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6
Sure, take the subgroups generated by $(12)$ and $(34)$ inside $S_4$. It's probably more interesting to ask that neither $H$ nor $K$ be normal in $HK$. For that case, consider subgroups where $|HK|$ is equal to $|G|$. You can always find $|HK|$ in terms of $|H|$ and $|K|$.
– Steve D
Jan 19 '18 at 18:39
@Steve D: Great! Thank you very much.
– Moritz
Jan 19 '18 at 18:53
3
For a specific example of this take $G=S_n$, $H$ the cyclic subgroup $langle (1,2,3,ldots,n) rangle$, and $K cong S_{n-1}$ a point stabilizer.
– Derek Holt
Jan 19 '18 at 19:54
@Derek Holt: A fine example! Thank you.
– Moritz
Jan 20 '18 at 16:45