Is $G/langle xrangle cap langle x rangle=e$ given the following subgroups have order 11 and 7
$begingroup$
I was trying to prove a question in group theory and if I assumed that $G/langle xrangle cap langle x rangle=e$ it will solve my question. I believe this is true because I know that the order is as follows, $|G/langle xrangle|=11$ and $|langle x rangle|=7$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $G/langle xrangle$ is $langle x rangle$ so I am unsure if I am making a big mistake.
group-theory finite-groups
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
$endgroup$
add a comment |
$begingroup$
I was trying to prove a question in group theory and if I assumed that $G/langle xrangle cap langle x rangle=e$ it will solve my question. I believe this is true because I know that the order is as follows, $|G/langle xrangle|=11$ and $|langle x rangle|=7$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $G/langle xrangle$ is $langle x rangle$ so I am unsure if I am making a big mistake.
group-theory finite-groups
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
$endgroup$
1
$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
1
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46
add a comment |
$begingroup$
I was trying to prove a question in group theory and if I assumed that $G/langle xrangle cap langle x rangle=e$ it will solve my question. I believe this is true because I know that the order is as follows, $|G/langle xrangle|=11$ and $|langle x rangle|=7$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $G/langle xrangle$ is $langle x rangle$ so I am unsure if I am making a big mistake.
group-theory finite-groups
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
$endgroup$
I was trying to prove a question in group theory and if I assumed that $G/langle xrangle cap langle x rangle=e$ it will solve my question. I believe this is true because I know that the order is as follows, $|G/langle xrangle|=11$ and $|langle x rangle|=7$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $G/langle xrangle$ is $langle x rangle$ so I am unsure if I am making a big mistake.
group-theory finite-groups
group-theory finite-groups
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
edited Dec 20 '18 at 18:46
Arturo Magidin
266k34590920
266k34590920
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
asked Dec 18 '18 at 19:53
AppleguysfriendAppleguysfriend
643
643
1
$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
1
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46
add a comment |
1
$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
1
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46
1
1
$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
1
1
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually $G / langle x rangle cap langle x rangle = emptyset$ since $G / langle x rangle$ has elements that are cosets and $langle x rangle$ contains elements of $G$.
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
$endgroup$
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
add a comment |
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1 Answer
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$begingroup$
Actually $G / langle x rangle cap langle x rangle = emptyset$ since $G / langle x rangle$ has elements that are cosets and $langle x rangle$ contains elements of $G$.
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
$endgroup$
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
add a comment |
$begingroup$
Actually $G / langle x rangle cap langle x rangle = emptyset$ since $G / langle x rangle$ has elements that are cosets and $langle x rangle$ contains elements of $G$.
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
$endgroup$
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
add a comment |
$begingroup$
Actually $G / langle x rangle cap langle x rangle = emptyset$ since $G / langle x rangle$ has elements that are cosets and $langle x rangle$ contains elements of $G$.
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
$endgroup$
Actually $G / langle x rangle cap langle x rangle = emptyset$ since $G / langle x rangle$ has elements that are cosets and $langle x rangle$ contains elements of $G$.
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
answered Dec 18 '18 at 20:00
PerturbativePerturbative
4,51121554
4,51121554
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
add a comment |
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
1
1
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
$begingroup$
Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ...
$endgroup$
– Henning Makholm
Dec 20 '18 at 18:52
add a comment |
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$begingroup$
$G/langle xrangle$ is a group whose underlying set is a set of cosets of $G$; $langle xrangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense.
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:15
1
$begingroup$
@C Monsour: Your “correction” cannot be what was intended; for $langle xrangle$ is precisely the identity element of $/langle xrangle$; So you are basically rewriting it to ask if $G/langle xrangle cap {e_{G/langle xrangle}} = {e_{G/langle xrangle}}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:45
$begingroup$
I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be.
$endgroup$
– Arturo Magidin
Dec 20 '18 at 18:46