Is $G/langle xrangle cap langle x rangle=e$ given the following subgroups have order 11 and 7












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 18:46









Arturo Magidin

266k34590920




266k34590920










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














  • 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










1 Answer
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1












$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$









  • 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












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$









  • 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
















1












$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$









  • 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














1












1








1





$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$



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 $G$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















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