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Understanding Converse of Lagrange's theorem!
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Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)
The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)
If my statement is $pto q$ then converse is $qto p$.
i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $pto q$ then converse is $qto p$
logic finite-groups
asked Aug 12 at 8:59
Cloud JR
543415
add a comment |Â
up vote
2
down vote
favorite
Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)
The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)
If my statement is $pto q$ then converse is $qto p$.
i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $pto q$ then converse is $qto p$
logic finite-groups
asked Aug 12 at 8:59
Cloud JR
543415
(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)
The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)
If my statement is $pto q$ then converse is $qto p$.
i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $pto q$ then converse is $qto p$
logic finite-groups
asked Aug 12 at 8:59
Cloud JR
543415
Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)
The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)
If my statement is $pto q$ then converse is $qto p$.
i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $pto q$ then converse is $qto p$
logic finite-groups
asked Aug 12 at 8:59
Cloud JR
543415
edited Aug 12 at 9:10
asked Aug 12 at 8:59
Cloud JR
543415
asked Aug 12 at 8:59
Cloud JR
543415
asked Aug 12 at 8:59
Cloud JR
543415
543415
(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05
add a comment |Â
(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05
(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Actually, you need to write the Lagrange theorem in $p rightarrow q$ form. So, the Lagrange theorem is actually,
If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.
Here $p:$ $H$ is a subgroup of $G$
$q;$ Order of $H$ divides order of $G$.
Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.
So, the converse will be
If a number $m$ divides order of $G$, then there is a subgroup of order $m$.
I hope this gives you insights.
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
 |Â
show 2 more comments
up vote
0
down vote
What is exactly your question?
Do you want a counter example of the converse of Lagrange's theorem?
Such a counter example is follwing:
The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.
answered Aug 12 at 9:01
mathlettuce
225
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
add a comment |Â
up vote
-2
down vote
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!
answered Aug 12 at 9:02
Anik Bhowmick
480417
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Actually, you need to write the Lagrange theorem in $p rightarrow q$ form. So, the Lagrange theorem is actually,
If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.
Here $p:$ $H$ is a subgroup of $G$
$q;$ Order of $H$ divides order of $G$.
Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.
So, the converse will be
If a number $m$ divides order of $G$, then there is a subgroup of order $m$.
I hope this gives you insights.
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
 |Â
show 2 more comments
up vote
5
down vote
accepted
Actually, you need to write the Lagrange theorem in $p rightarrow q$ form. So, the Lagrange theorem is actually,
If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.
Here $p:$ $H$ is a subgroup of $G$
$q;$ Order of $H$ divides order of $G$.
Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.
So, the converse will be
If a number $m$ divides order of $G$, then there is a subgroup of order $m$.
I hope this gives you insights.
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
 |Â
show 2 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Actually, you need to write the Lagrange theorem in $p rightarrow q$ form. So, the Lagrange theorem is actually,
If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.
Here $p:$ $H$ is a subgroup of $G$
$q;$ Order of $H$ divides order of $G$.
Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.
So, the converse will be
If a number $m$ divides order of $G$, then there is a subgroup of order $m$.
I hope this gives you insights.
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
Actually, you need to write the Lagrange theorem in $p rightarrow q$ form. So, the Lagrange theorem is actually,
If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.
Here $p:$ $H$ is a subgroup of $G$
$q;$ Order of $H$ divides order of $G$.
Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.
So, the converse will be
If a number $m$ divides order of $G$, then there is a subgroup of order $m$.
I hope this gives you insights.
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
answered Aug 12 at 9:04
Aniruddha Deshmukh
663417
663417
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
 |Â
show 2 more comments
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
It seems that the OP gets trouble in understanding a logical statement.
– mrs
Aug 12 at 9:07
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
@ResidentDementor yes you are correct
– Cloud JR
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra".
– Aniruddha Deshmukh
Aug 12 at 9:08
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
well i change the tags.. thanks for your suggestion
– Cloud JR
Aug 12 at 9:10
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct?
– Cloud JR
Aug 12 at 9:13
 |Â
show 2 more comments
up vote
0
down vote
What is exactly your question?
Do you want a counter example of the converse of Lagrange's theorem?
Such a counter example is follwing:
The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.
answered Aug 12 at 9:01
mathlettuce
225
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
add a comment |Â
up vote
0
down vote
What is exactly your question?
Do you want a counter example of the converse of Lagrange's theorem?
Such a counter example is follwing:
The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.
answered Aug 12 at 9:01
mathlettuce
225
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What is exactly your question?
Do you want a counter example of the converse of Lagrange's theorem?
Such a counter example is follwing:
The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.
answered Aug 12 at 9:01
mathlettuce
225
What is exactly your question?
Do you want a counter example of the converse of Lagrange's theorem?
Such a counter example is follwing:
The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.
answered Aug 12 at 9:01
mathlettuce
225
edited Aug 12 at 9:03
answered Aug 12 at 9:01
mathlettuce
225
answered Aug 12 at 9:01
mathlettuce
225
answered Aug 12 at 9:01
mathlettuce
225
225
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
add a comment |Â
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
Better,you made this as comment
– mrs
Aug 12 at 9:02
Better,you made this as comment
– mrs
Aug 12 at 9:02
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so.
– mathlettuce
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
add a comment |Â
up vote
-2
down vote
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!
answered Aug 12 at 9:02
Anik Bhowmick
480417
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
add a comment |Â
up vote
-2
down vote
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!
answered Aug 12 at 9:02
Anik Bhowmick
480417
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!
answered Aug 12 at 9:02
Anik Bhowmick
480417
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!
answered Aug 12 at 9:02
Anik Bhowmick
480417
edited Aug 12 at 9:21
answered Aug 12 at 9:02
Anik Bhowmick
480417
answered Aug 12 at 9:02
Anik Bhowmick
480417
answered Aug 12 at 9:02
Anik Bhowmick
480417
480417
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
add a comment |Â
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
2
2
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
The order of $A_4$ is $12$
– leibnewtz
Aug 12 at 9:20
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
Yes, you're right @leibnewtz. My mistake !!
– Anik Bhowmick
Aug 12 at 9:21
add a comment |Â
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(I'm sorry if this question is uninteresting But please help!)
– Cloud JR
Aug 12 at 8:59
See math.stackexchange.com/a/41762/8581
– mrs
Aug 12 at 9:03
I don't want counter example! i want to know how (2) is converse of (1)
– Cloud JR
Aug 12 at 9:05
I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct!
– Aniruddha Deshmukh
Aug 12 at 9:05