Does $HK = G$ imply that $KH = G$ when $H,K$ aren't subgroups?
$begingroup$
Let $G$ be a finite group and let $H$ and $K$ be two "subsets" of $G$ such that $HK = G$. Does that imply that $KH = G$?
Since $H$ and $K$ are not subgroups we cannot use the formula $o(HK) = frac{o(H)o(K)}{o(Hcap K)}$.
I believe this is false and I am looking for counterexamples.
Thank you.
group-theory finite-groups
edited Dec 17 '18 at 5:18
the_fox
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group and let $H$ and $K$ be two "subsets" of $G$ such that $HK = G$. Does that imply that $KH = G$?
Since $H$ and $K$ are not subgroups we cannot use the formula $o(HK) = frac{o(H)o(K)}{o(Hcap K)}$.
I believe this is false and I am looking for counterexamples.
Thank you.
group-theory finite-groups
edited Dec 17 '18 at 5:18
the_fox
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
$endgroup$
$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45
add a comment |
$begingroup$
Let $G$ be a finite group and let $H$ and $K$ be two "subsets" of $G$ such that $HK = G$. Does that imply that $KH = G$?
Since $H$ and $K$ are not subgroups we cannot use the formula $o(HK) = frac{o(H)o(K)}{o(Hcap K)}$.
I believe this is false and I am looking for counterexamples.
Thank you.
group-theory finite-groups
edited Dec 17 '18 at 5:18
the_fox
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
$endgroup$
Let $G$ be a finite group and let $H$ and $K$ be two "subsets" of $G$ such that $HK = G$. Does that imply that $KH = G$?
Since $H$ and $K$ are not subgroups we cannot use the formula $o(HK) = frac{o(H)o(K)}{o(Hcap K)}$.
I believe this is false and I am looking for counterexamples.
Thank you.
group-theory finite-groups
group-theory finite-groups
edited Dec 17 '18 at 5:18
the_fox
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
edited Dec 17 '18 at 5:18
the_fox
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
edited Dec 17 '18 at 5:18
the_fox
2,90031538
edited Dec 17 '18 at 5:18
the_fox
2,90031538
edited Dec 17 '18 at 5:18
the_fox
2,90031538
2,90031538
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
asked Dec 17 '18 at 3:33
GA316GA316
2,6891232
2,6891232
$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45
add a comment |
$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45
$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and
$$H= { e, (1,2) } \
K={ e, (1,3), (1,2,3) }$$
Then
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and
$$H= { e, (1,2) } \
K={ e, (1,3), (1,2,3) }$$
Then
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
add a comment |
$begingroup$
Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and
$$H= { e, (1,2) } \
K={ e, (1,3), (1,2,3) }$$
Then
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
add a comment |
$begingroup$
Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and
$$H= { e, (1,2) } \
K={ e, (1,3), (1,2,3) }$$
Then
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
$endgroup$
Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and
$$H= { e, (1,2) } \
K={ e, (1,3), (1,2,3) }$$
Then
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
answered Dec 17 '18 at 4:05
N. S.N. S.
104k7114209
104k7114209
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
add a comment |
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
$begingroup$
Thank you N.S. This example is simple and nice.
$endgroup$
– GA316
Dec 17 '18 at 5:35
add a comment |
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$begingroup$
Conjugation perhaps?
$endgroup$
– Robert Wolfe
Dec 17 '18 at 3:44
$begingroup$
If $H$ and $K$ are not subgroups but are closed under inverses, then it's true. So you need one of them to fail to be closed under inverses.
$endgroup$
– Matt Samuel
Dec 17 '18 at 3:45