Mixing
Prove $Hcap K$ Abelian.
Let $H$ and $K$ be subgroups of a group with $|H|=24$ and $|K|=20$. Prove that $Hcap K$ Abelian.
I am unsure where to start with this proof. Any help is appreciated.
abstract-algebra
asked Nov 28 '17 at 15:43
rover2
769212
add a comment |
Let $H$ and $K$ be subgroups of a group with $|H|=24$ and $|K|=20$. Prove that $Hcap K$ Abelian.
I am unsure where to start with this proof. Any help is appreciated.
abstract-algebra
asked Nov 28 '17 at 15:43
rover2
769212
4
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47
add a comment |
Let $H$ and $K$ be subgroups of a group with $|H|=24$ and $|K|=20$. Prove that $Hcap K$ Abelian.
I am unsure where to start with this proof. Any help is appreciated.
abstract-algebra
asked Nov 28 '17 at 15:43
rover2
769212
Let $H$ and $K$ be subgroups of a group with $|H|=24$ and $|K|=20$. Prove that $Hcap K$ Abelian.
I am unsure where to start with this proof. Any help is appreciated.
abstract-algebra
abstract-algebra
asked Nov 28 '17 at 15:43
rover2
769212
asked Nov 28 '17 at 15:43
rover2
769212
asked Nov 28 '17 at 15:43
rover2
769212
asked Nov 28 '17 at 15:43
rover2
769212
asked Nov 28 '17 at 15:43
rover2
769212
769212
4
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47
add a comment |
4
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47
4
4
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47
add a comment |
2 Answers
2
active
oldest
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$Hcap K$ is a group being the intersection of two groups. Also it is a subgroup of both, so its order must divide both $20$ and $24$, hence the order of $Hcap K$ can be atmost $4$. Let $x$ and $y$ be two elements of $Hcap K$. If $x$ and $y$ doesn't commute then show that $e,x,y,xy,yx$ are $5$ distinct elements of $Hcap K$, which leads to a contradiction. So any group having order $leq 4$ must be abelian.
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
add a comment |
We know that $Hcap K$ subgroup for subgroups $H$ and $K$.Therefore,
$|Hcap K|$ could be $1, 2, 4$ by Lagrance Theorem. We know also that if $G$ is a group with $|G|leq 5$, then $G$ is abealian. In particular,
$textbf{ Case 1)}$ If $|Hcap K|=1$, then $Hcap K={1}$.
$textbf{ Case 2)}$ if $|Hcap K|=2$, then we know that every prime order is cyclic and every cyclic group is abelian.
$textbf{ Case 3)}$ If $|Hcap K|=4$, then it is isomorphic to either $mathbb{Z_4}$ or $4$-group $V$ which are abelian.
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
add a comment |
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2 Answers
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$Hcap K$ is a group being the intersection of two groups. Also it is a subgroup of both, so its order must divide both $20$ and $24$, hence the order of $Hcap K$ can be atmost $4$. Let $x$ and $y$ be two elements of $Hcap K$. If $x$ and $y$ doesn't commute then show that $e,x,y,xy,yx$ are $5$ distinct elements of $Hcap K$, which leads to a contradiction. So any group having order $leq 4$ must be abelian.
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
add a comment |
$Hcap K$ is a group being the intersection of two groups. Also it is a subgroup of both, so its order must divide both $20$ and $24$, hence the order of $Hcap K$ can be atmost $4$. Let $x$ and $y$ be two elements of $Hcap K$. If $x$ and $y$ doesn't commute then show that $e,x,y,xy,yx$ are $5$ distinct elements of $Hcap K$, which leads to a contradiction. So any group having order $leq 4$ must be abelian.
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
add a comment |
$Hcap K$ is a group being the intersection of two groups. Also it is a subgroup of both, so its order must divide both $20$ and $24$, hence the order of $Hcap K$ can be atmost $4$. Let $x$ and $y$ be two elements of $Hcap K$. If $x$ and $y$ doesn't commute then show that $e,x,y,xy,yx$ are $5$ distinct elements of $Hcap K$, which leads to a contradiction. So any group having order $leq 4$ must be abelian.
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
$Hcap K$ is a group being the intersection of two groups. Also it is a subgroup of both, so its order must divide both $20$ and $24$, hence the order of $Hcap K$ can be atmost $4$. Let $x$ and $y$ be two elements of $Hcap K$. If $x$ and $y$ doesn't commute then show that $e,x,y,xy,yx$ are $5$ distinct elements of $Hcap K$, which leads to a contradiction. So any group having order $leq 4$ must be abelian.
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
answered Nov 28 '17 at 15:48
Abishanka Saha
7,79811022
7,79811022
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
add a comment |
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Just to add that, if we can assume the OP already knows that all groups of order 1, 2 and 4 are Abelian, the proof gets even simpler.
– user491874
Nov 28 '17 at 15:55
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
Yes, this is quite trivial. We showed this our first course in group theory.
– Math_QED
Nov 28 '17 at 15:58
add a comment |
We know that $Hcap K$ subgroup for subgroups $H$ and $K$.Therefore,
$|Hcap K|$ could be $1, 2, 4$ by Lagrance Theorem. We know also that if $G$ is a group with $|G|leq 5$, then $G$ is abealian. In particular,
$textbf{ Case 1)}$ If $|Hcap K|=1$, then $Hcap K={1}$.
$textbf{ Case 2)}$ if $|Hcap K|=2$, then we know that every prime order is cyclic and every cyclic group is abelian.
$textbf{ Case 3)}$ If $|Hcap K|=4$, then it is isomorphic to either $mathbb{Z_4}$ or $4$-group $V$ which are abelian.
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
add a comment |
We know that $Hcap K$ subgroup for subgroups $H$ and $K$.Therefore,
$|Hcap K|$ could be $1, 2, 4$ by Lagrance Theorem. We know also that if $G$ is a group with $|G|leq 5$, then $G$ is abealian. In particular,
$textbf{ Case 1)}$ If $|Hcap K|=1$, then $Hcap K={1}$.
$textbf{ Case 2)}$ if $|Hcap K|=2$, then we know that every prime order is cyclic and every cyclic group is abelian.
$textbf{ Case 3)}$ If $|Hcap K|=4$, then it is isomorphic to either $mathbb{Z_4}$ or $4$-group $V$ which are abelian.
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
add a comment |
We know that $Hcap K$ subgroup for subgroups $H$ and $K$.Therefore,
$|Hcap K|$ could be $1, 2, 4$ by Lagrance Theorem. We know also that if $G$ is a group with $|G|leq 5$, then $G$ is abealian. In particular,
$textbf{ Case 1)}$ If $|Hcap K|=1$, then $Hcap K={1}$.
$textbf{ Case 2)}$ if $|Hcap K|=2$, then we know that every prime order is cyclic and every cyclic group is abelian.
$textbf{ Case 3)}$ If $|Hcap K|=4$, then it is isomorphic to either $mathbb{Z_4}$ or $4$-group $V$ which are abelian.
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
We know that $Hcap K$ subgroup for subgroups $H$ and $K$.Therefore,
$|Hcap K|$ could be $1, 2, 4$ by Lagrance Theorem. We know also that if $G$ is a group with $|G|leq 5$, then $G$ is abealian. In particular,
$textbf{ Case 1)}$ If $|Hcap K|=1$, then $Hcap K={1}$.
$textbf{ Case 2)}$ if $|Hcap K|=2$, then we know that every prime order is cyclic and every cyclic group is abelian.
$textbf{ Case 3)}$ If $|Hcap K|=4$, then it is isomorphic to either $mathbb{Z_4}$ or $4$-group $V$ which are abelian.
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
edited Nov 21 '18 at 7:53
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
answered Nov 28 '17 at 16:02
1ENİGMA1
958416
958416
add a comment |
add a comment |
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4
$Hcap K$ is a subgroup of both $H$ and $K$, therefore its order must...
– user228113
Nov 28 '17 at 15:44
its order must divide $|HK|$?? or both $|H|$ and $|K|$?
– rover2
Nov 28 '17 at 15:47