Does $HK = G$ imply that $KH = G$ when $H,K$ aren't subgroups?












6












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















6












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















6












6








6


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 5:18









the_fox

2,90031538




2,90031538










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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you N.S. This example is simple and nice.
    $endgroup$
    – GA316
    Dec 17 '18 at 5:35











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you N.S. This example is simple and nice.
    $endgroup$
    – GA316
    Dec 17 '18 at 5:35
















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you N.S. This example is simple and nice.
    $endgroup$
    – GA316
    Dec 17 '18 at 5:35














11












11








11





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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