Inverting without actual inverse?
$begingroup$
I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following:
For which I guess the problem lies when we try to take an element $hkin HK$, it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
My problem is the following, taking the definition of subgroup, we have:
- If $a,bin H$, then $ab^{-1}in H$.
Then, all the elements $kcdot1=k$ and $1cdot h=h$ are in $HK$, that is: We have all the elements of both $H,K$ in $HK$. By the definition, suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before but $(hk)^{-1}$ is not actually in $HK$.
abstract-algebra group-theory
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
add a comment |
$begingroup$
I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following:
For which I guess the problem lies when we try to take an element $hkin HK$, it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
My problem is the following, taking the definition of subgroup, we have:
- If $a,bin H$, then $ab^{-1}in H$.
Then, all the elements $kcdot1=k$ and $1cdot h=h$ are in $HK$, that is: We have all the elements of both $H,K$ in $HK$. By the definition, suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before but $(hk)^{-1}$ is not actually in $HK$.
abstract-algebra group-theory
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
1
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
1
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18
add a comment |
$begingroup$
I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following:
For which I guess the problem lies when we try to take an element $hkin HK$, it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
My problem is the following, taking the definition of subgroup, we have:
- If $a,bin H$, then $ab^{-1}in H$.
Then, all the elements $kcdot1=k$ and $1cdot h=h$ are in $HK$, that is: We have all the elements of both $H,K$ in $HK$. By the definition, suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before but $(hk)^{-1}$ is not actually in $HK$.
abstract-algebra group-theory
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following:
For which I guess the problem lies when we try to take an element $hkin HK$, it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
My problem is the following, taking the definition of subgroup, we have:
- If $a,bin H$, then $ab^{-1}in H$.
Then, all the elements $kcdot1=k$ and $1cdot h=h$ are in $HK$, that is: We have all the elements of both $H,K$ in $HK$. By the definition, suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before but $(hk)^{-1}$ is not actually in $HK$.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
edited Jan 2 at 8:59
Billy Rubina
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 2 at 8:50
Billy RubinaBilly Rubina
10.4k1458134
10.4k1458134
1
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
1
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18
add a comment |
1
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
1
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18
1
1
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
1
1
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18
add a comment |
2 Answers
2
active
oldest
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$begingroup$
$newcommand{Span}[1]{leftlangle #1 rightrangle}$Allow me first of all to question your statement
when we try to take an element $hkin HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = Span{(1 2)}$, $K = Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) ne (1 3) = k^{-1} h^{-1} in H K = S_{3}$.
As to your second statement
suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
$endgroup$
add a comment |
$begingroup$
Assume $HK = KH$. To show that $HK$ is a subgroup, consider $g_1= h_1k_1, g_2 = h_2k_2$ in $HK$. We have $(k_1k_2^{-1})h_2^{-1} in KH = HK$, hence $(k_1k_2^{-1})h_2^{-1} = hk$ with $h in H, k in K$. Then $g_1g_2^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1hk in HK$.
If you consider $hk in HK$, then you see even easier $(hk)^{-1} = k^{-1}h^{-1} in KH = HK$.
Conversely, assume that $HK$ is a subgroup of $G$.
(a) $KH subset HK$.
Let $g in KH$. Write $g = kh$ with $h in H, k in K$. Then $g^{-1} = h^{-1}k^{-1} in HK$, hence $g in HK$.
(b) $HK subset KH$.
Let $g in HK$. Then also $g^{-1} in HK$, hence $g^{-1} = hk$ with $h in H, k in K$. This shows $g = k^{-1}h^{-1} in KH$.
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$newcommand{Span}[1]{leftlangle #1 rightrangle}$Allow me first of all to question your statement
when we try to take an element $hkin HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = Span{(1 2)}$, $K = Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) ne (1 3) = k^{-1} h^{-1} in H K = S_{3}$.
As to your second statement
suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
$endgroup$
add a comment |
$begingroup$
$newcommand{Span}[1]{leftlangle #1 rightrangle}$Allow me first of all to question your statement
when we try to take an element $hkin HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = Span{(1 2)}$, $K = Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) ne (1 3) = k^{-1} h^{-1} in H K = S_{3}$.
As to your second statement
suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
$endgroup$
add a comment |
$begingroup$
$newcommand{Span}[1]{leftlangle #1 rightrangle}$Allow me first of all to question your statement
when we try to take an element $hkin HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = Span{(1 2)}$, $K = Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) ne (1 3) = k^{-1} h^{-1} in H K = S_{3}$.
As to your second statement
suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
$endgroup$
$newcommand{Span}[1]{leftlangle #1 rightrangle}$Allow me first of all to question your statement
when we try to take an element $hkin HK$, its inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$.
This is not true. Take for instance $G = S_{3}$, $H = Span{(1 2)}$, $K = Span{(1 2 3)}$, $h = (1 2)$, $k = (1 2 3)$. Then $h^{-1} k^{-1} = (2 3) ne (1 3) = k^{-1} h^{-1} in H K = S_{3}$.
As to your second statement
suppose we have $hkin HK$, then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$. Isn't this very similar to having the inverse of $hk$ in $HK$?
I fail to see the point. Sure, $h k$ has an inverse $k^{-1} h^{-1}$ in $G$, and that's what you are checking. But to show that $H K$ is a subgroup you will have to show that (under suitable conditions) this inverse lies precisely in $H K$.
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
answered Jan 2 at 9:01
Andreas CarantiAndreas Caranti
56.4k34395
56.4k34395
add a comment |
add a comment |
$begingroup$
Assume $HK = KH$. To show that $HK$ is a subgroup, consider $g_1= h_1k_1, g_2 = h_2k_2$ in $HK$. We have $(k_1k_2^{-1})h_2^{-1} in KH = HK$, hence $(k_1k_2^{-1})h_2^{-1} = hk$ with $h in H, k in K$. Then $g_1g_2^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1hk in HK$.
If you consider $hk in HK$, then you see even easier $(hk)^{-1} = k^{-1}h^{-1} in KH = HK$.
Conversely, assume that $HK$ is a subgroup of $G$.
(a) $KH subset HK$.
Let $g in KH$. Write $g = kh$ with $h in H, k in K$. Then $g^{-1} = h^{-1}k^{-1} in HK$, hence $g in HK$.
(b) $HK subset KH$.
Let $g in HK$. Then also $g^{-1} in HK$, hence $g^{-1} = hk$ with $h in H, k in K$. This shows $g = k^{-1}h^{-1} in KH$.
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
$endgroup$
add a comment |
$begingroup$
Assume $HK = KH$. To show that $HK$ is a subgroup, consider $g_1= h_1k_1, g_2 = h_2k_2$ in $HK$. We have $(k_1k_2^{-1})h_2^{-1} in KH = HK$, hence $(k_1k_2^{-1})h_2^{-1} = hk$ with $h in H, k in K$. Then $g_1g_2^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1hk in HK$.
If you consider $hk in HK$, then you see even easier $(hk)^{-1} = k^{-1}h^{-1} in KH = HK$.
Conversely, assume that $HK$ is a subgroup of $G$.
(a) $KH subset HK$.
Let $g in KH$. Write $g = kh$ with $h in H, k in K$. Then $g^{-1} = h^{-1}k^{-1} in HK$, hence $g in HK$.
(b) $HK subset KH$.
Let $g in HK$. Then also $g^{-1} in HK$, hence $g^{-1} = hk$ with $h in H, k in K$. This shows $g = k^{-1}h^{-1} in KH$.
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
$endgroup$
add a comment |
$begingroup$
Assume $HK = KH$. To show that $HK$ is a subgroup, consider $g_1= h_1k_1, g_2 = h_2k_2$ in $HK$. We have $(k_1k_2^{-1})h_2^{-1} in KH = HK$, hence $(k_1k_2^{-1})h_2^{-1} = hk$ with $h in H, k in K$. Then $g_1g_2^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1hk in HK$.
If you consider $hk in HK$, then you see even easier $(hk)^{-1} = k^{-1}h^{-1} in KH = HK$.
Conversely, assume that $HK$ is a subgroup of $G$.
(a) $KH subset HK$.
Let $g in KH$. Write $g = kh$ with $h in H, k in K$. Then $g^{-1} = h^{-1}k^{-1} in HK$, hence $g in HK$.
(b) $HK subset KH$.
Let $g in HK$. Then also $g^{-1} in HK$, hence $g^{-1} = hk$ with $h in H, k in K$. This shows $g = k^{-1}h^{-1} in KH$.
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
$endgroup$
Assume $HK = KH$. To show that $HK$ is a subgroup, consider $g_1= h_1k_1, g_2 = h_2k_2$ in $HK$. We have $(k_1k_2^{-1})h_2^{-1} in KH = HK$, hence $(k_1k_2^{-1})h_2^{-1} = hk$ with $h in H, k in K$. Then $g_1g_2^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1hk in HK$.
If you consider $hk in HK$, then you see even easier $(hk)^{-1} = k^{-1}h^{-1} in KH = HK$.
Conversely, assume that $HK$ is a subgroup of $G$.
(a) $KH subset HK$.
Let $g in KH$. Write $g = kh$ with $h in H, k in K$. Then $g^{-1} = h^{-1}k^{-1} in HK$, hence $g in HK$.
(b) $HK subset KH$.
Let $g in HK$. Then also $g^{-1} in HK$, hence $g^{-1} = hk$ with $h in H, k in K$. This shows $g = k^{-1}h^{-1} in KH$.
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
edited Jan 2 at 10:24
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
answered Jan 2 at 9:20
Paul FrostPaul Frost
9,9653932
9,9653932
add a comment |
add a comment |
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1
$begingroup$
It is not clear to me what you are asking. I guess you want to prove that $HK le G$ implies $HK = KH$. The converse is fairly obvious.
$endgroup$
– Paul Frost
Jan 2 at 10:08
1
$begingroup$
I think your confusion stems from the way you're looking at this. Notice that $(hk)(k^{-1})$ must be in $HK$ trivially since it's $h$ same for $(hk)(k^{-1})(h^{-1})$ since that is just $1$. But just because $abin HK$ that says nothing about $a$ or $b$ being in $HK$ unless you know $HK$ is a group. Even if you know $ain HK$ and $abin HK$ you only get $b$ in $HK$ if $HK$ is a group.
$endgroup$
– DRF
Jan 2 at 10:18