Mixing
Converse of Theorem Proof/Counterexample
$begingroup$
Theorem: If $G$ is an abelian group such that $Hleq G$, then $H$ is an abelian subgroup.
I've proved this theorem here:
Proof: Suppose $G$ is abelian. Let $a,bin G$. Then $ab=ba$ as G is abelian and commutative properties are inherited. Since $Hleq G$, then for all $a,bin H, ab=ba$ also. Thus, $H$ is abelian.
I am new and still learning how to construct proofs -- please be kind!
The question: Is the converse of the theorem true?
I am having a hard time understanding exactly what the text means by just the converse of the theorem. Are they asking "If $G$ is not an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup? Or, are they asking, "If $G$ is an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup?
If $G$ is an abelian group and if $Hleq G$, then can $H$ be an abelian subgroup? Well, isn't this kind of trivial. Suppose $G$ is abelian. Then laws of commutative properties and other group properties follow; that is, there exist an inverse, identity, and $H$ would also be closed under the binary operations *, where the operation * is associative. Then, wouldn't all elements of $H$ also be abelian since they inherit all these properties from $G$?
abstract-algebra abelian-groups
asked Feb 2 at 22:47
RyanRyan
1899
$endgroup$
add a comment |
$begingroup$
Theorem: If $G$ is an abelian group such that $Hleq G$, then $H$ is an abelian subgroup.
I've proved this theorem here:
Proof: Suppose $G$ is abelian. Let $a,bin G$. Then $ab=ba$ as G is abelian and commutative properties are inherited. Since $Hleq G$, then for all $a,bin H, ab=ba$ also. Thus, $H$ is abelian.
I am new and still learning how to construct proofs -- please be kind!
The question: Is the converse of the theorem true?
I am having a hard time understanding exactly what the text means by just the converse of the theorem. Are they asking "If $G$ is not an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup? Or, are they asking, "If $G$ is an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup?
If $G$ is an abelian group and if $Hleq G$, then can $H$ be an abelian subgroup? Well, isn't this kind of trivial. Suppose $G$ is abelian. Then laws of commutative properties and other group properties follow; that is, there exist an inverse, identity, and $H$ would also be closed under the binary operations *, where the operation * is associative. Then, wouldn't all elements of $H$ also be abelian since they inherit all these properties from $G$?
abstract-algebra abelian-groups
asked Feb 2 at 22:47
RyanRyan
1899
$endgroup$
$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03
add a comment |
$begingroup$
Theorem: If $G$ is an abelian group such that $Hleq G$, then $H$ is an abelian subgroup.
I've proved this theorem here:
Proof: Suppose $G$ is abelian. Let $a,bin G$. Then $ab=ba$ as G is abelian and commutative properties are inherited. Since $Hleq G$, then for all $a,bin H, ab=ba$ also. Thus, $H$ is abelian.
I am new and still learning how to construct proofs -- please be kind!
The question: Is the converse of the theorem true?
I am having a hard time understanding exactly what the text means by just the converse of the theorem. Are they asking "If $G$ is not an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup? Or, are they asking, "If $G$ is an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup?
If $G$ is an abelian group and if $Hleq G$, then can $H$ be an abelian subgroup? Well, isn't this kind of trivial. Suppose $G$ is abelian. Then laws of commutative properties and other group properties follow; that is, there exist an inverse, identity, and $H$ would also be closed under the binary operations *, where the operation * is associative. Then, wouldn't all elements of $H$ also be abelian since they inherit all these properties from $G$?
abstract-algebra abelian-groups
asked Feb 2 at 22:47
RyanRyan
1899
$endgroup$
Theorem: If $G$ is an abelian group such that $Hleq G$, then $H$ is an abelian subgroup.
I've proved this theorem here:
Proof: Suppose $G$ is abelian. Let $a,bin G$. Then $ab=ba$ as G is abelian and commutative properties are inherited. Since $Hleq G$, then for all $a,bin H, ab=ba$ also. Thus, $H$ is abelian.
I am new and still learning how to construct proofs -- please be kind!
The question: Is the converse of the theorem true?
I am having a hard time understanding exactly what the text means by just the converse of the theorem. Are they asking "If $G$ is not an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup? Or, are they asking, "If $G$ is an abelian group such that $Hleq G$, then $H$ is not an abelian subgroup?
If $G$ is an abelian group and if $Hleq G$, then can $H$ be an abelian subgroup? Well, isn't this kind of trivial. Suppose $G$ is abelian. Then laws of commutative properties and other group properties follow; that is, there exist an inverse, identity, and $H$ would also be closed under the binary operations *, where the operation * is associative. Then, wouldn't all elements of $H$ also be abelian since they inherit all these properties from $G$?
abstract-algebra abelian-groups
abstract-algebra abelian-groups
asked Feb 2 at 22:47
RyanRyan
1899
asked Feb 2 at 22:47
RyanRyan
1899
asked Feb 2 at 22:47
RyanRyan
1899
asked Feb 2 at 22:47
RyanRyan
1899
asked Feb 2 at 22:47
RyanRyan
1899
1899
$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03
add a comment |
$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03
$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
$endgroup$
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
add a comment |
$begingroup$
The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.
The other possible interpretations appear to be trivial.
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
$endgroup$
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
add a comment |
$begingroup$
The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
$endgroup$
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
add a comment |
$begingroup$
The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
$endgroup$
The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
answered Feb 2 at 22:53
M. RinettiM. Rinetti
307
307
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
add a comment |
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
1
1
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
$begingroup$
Referring to the "converse" here is hopelessly ambiguous because it is unclear what is supposed to happen to the quantifiers and the background assumptions. It's anyone's guess what the author of the problem actually intended.
$endgroup$
– Eric Wofsey
Feb 2 at 22:55
3
3
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
For instance, another reasonable "converse" would be: if $G$ is a group such that every subgroup of $G$ is abelian, then $G$ is abelian. Or another one: if $G$ is an abelian group and $H$ is an abelian subgroup of $G$, then $Hleq G$. Your interpretation is probably the most likely one since it is the most interesting one I can think of but really it is a terribly written problem.
$endgroup$
– Eric Wofsey
Feb 2 at 23:01
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
$begingroup$
You're right, reading it better the text is indeed ambiguous. I thought the statement I wrote was the most natural possible converse, even if thinking better I know it is wrong. In particular it's likely that the author intended the first one you give, even if there's no reason to prefer either of them. Thank you for your pointing that out.
$endgroup$
– M. Rinetti
Feb 2 at 23:12
add a comment |
$begingroup$
The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.
The other possible interpretations appear to be trivial.
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.
The other possible interpretations appear to be trivial.
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.
The other possible interpretations appear to be trivial.
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
$endgroup$
The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.
The other possible interpretations appear to be trivial.
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
edited Feb 3 at 0:08
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 23:39
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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$begingroup$
Have you copied the statement of the Theorem exactly word-for-word? If so, it is poorly worded and it is very unclear what is intended to be the "converse".
$endgroup$
– Eric Wofsey
Feb 2 at 22:58
$begingroup$
Hi @EricWofsey. Yes, I have. It is poorly worded and I have just reached out to my professor about it.
$endgroup$
– Ryan
Feb 2 at 23:03