How to quotient modules in homology computation












2












$begingroup$


I was computing the homology group of a 2-complex with $mathbb{Z}$ coefficients. Doing so, I obtained
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
$$
I then tried to simplify the expression for $H_1$ in two different ways. Unfortunately, I obtained two different answers. So at most one (if any) can be correct. I was hoping someone could explain where I went wrong and why my step is incorrect.



First Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
= frac{langle a,b,c,d rangle}{langle a = -b, c=-drangle}
cong mathbb{Z}^2
$$



Second Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
cong frac{langle a,a+b,c,c+d rangle}{langle 2(a+b), c+d rangle} cong mathbb{Z}_2 times mathbb{Z}^2
$$






abstract-algebra group-theory algebraic-topology homology-cohomology







share|cite|improve this question









$endgroup$












  • $begingroup$
    That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
    $endgroup$
    – reuns
    Dec 13 '18 at 4:18






  • 1




    $begingroup$
    First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
    $endgroup$
    – jgon
    Dec 13 '18 at 4:19






  • 1




    $begingroup$
    @reuns So the second one is correct?
    $endgroup$
    – Quoka
    Dec 13 '18 at 4:22
















2












$begingroup$


I was computing the homology group of a 2-complex with $mathbb{Z}$ coefficients. Doing so, I obtained
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
$$
I then tried to simplify the expression for $H_1$ in two different ways. Unfortunately, I obtained two different answers. So at most one (if any) can be correct. I was hoping someone could explain where I went wrong and why my step is incorrect.



First Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
= frac{langle a,b,c,d rangle}{langle a = -b, c=-drangle}
cong mathbb{Z}^2
$$



Second Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
cong frac{langle a,a+b,c,c+d rangle}{langle 2(a+b), c+d rangle} cong mathbb{Z}_2 times mathbb{Z}^2
$$






abstract-algebra group-theory algebraic-topology homology-cohomology







share|cite|improve this question









$endgroup$












  • $begingroup$
    That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
    $endgroup$
    – reuns
    Dec 13 '18 at 4:18






  • 1




    $begingroup$
    First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
    $endgroup$
    – jgon
    Dec 13 '18 at 4:19






  • 1




    $begingroup$
    @reuns So the second one is correct?
    $endgroup$
    – Quoka
    Dec 13 '18 at 4:22














2












2








2





$begingroup$


I was computing the homology group of a 2-complex with $mathbb{Z}$ coefficients. Doing so, I obtained
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
$$
I then tried to simplify the expression for $H_1$ in two different ways. Unfortunately, I obtained two different answers. So at most one (if any) can be correct. I was hoping someone could explain where I went wrong and why my step is incorrect.



First Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
= frac{langle a,b,c,d rangle}{langle a = -b, c=-drangle}
cong mathbb{Z}^2
$$



Second Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
cong frac{langle a,a+b,c,c+d rangle}{langle 2(a+b), c+d rangle} cong mathbb{Z}_2 times mathbb{Z}^2
$$






abstract-algebra group-theory algebraic-topology homology-cohomology







share|cite|improve this question









$endgroup$




I was computing the homology group of a 2-complex with $mathbb{Z}$ coefficients. Doing so, I obtained
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
$$
I then tried to simplify the expression for $H_1$ in two different ways. Unfortunately, I obtained two different answers. So at most one (if any) can be correct. I was hoping someone could explain where I went wrong and why my step is incorrect.



First Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
= frac{langle a,b,c,d rangle}{langle a = -b, c=-drangle}
cong mathbb{Z}^2
$$



Second Method:
$$
H_1 = frac{langle a,b,c,d rangle}{langle 2a+2b, 2a+2b+c+drangle}
cong frac{langle a,a+b,c,c+d rangle}{langle 2(a+b), c+d rangle} cong mathbb{Z}_2 times mathbb{Z}^2
$$






abstract-algebra group-theory algebraic-topology homology-cohomology




abstract-algebra group-theory algebraic-topology homology-cohomology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 4:11









QuokaQuoka

1,264312




1,264312












  • $begingroup$
    That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
    $endgroup$
    – reuns
    Dec 13 '18 at 4:18






  • 1




    $begingroup$
    First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
    $endgroup$
    – jgon
    Dec 13 '18 at 4:19






  • 1




    $begingroup$
    @reuns So the second one is correct?
    $endgroup$
    – Quoka
    Dec 13 '18 at 4:22


















  • $begingroup$
    That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
    $endgroup$
    – reuns
    Dec 13 '18 at 4:18






  • 1




    $begingroup$
    First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
    $endgroup$
    – jgon
    Dec 13 '18 at 4:19






  • 1




    $begingroup$
    @reuns So the second one is correct?
    $endgroup$
    – Quoka
    Dec 13 '18 at 4:22
















$begingroup$
That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
$endgroup$
– reuns
Dec 13 '18 at 4:18




$begingroup$
That $2u = 0 implies u = 0$ means your group has no $2$-torsion. So doing that you have quotiented the correct group by its $2$-torsion.
$endgroup$
– reuns
Dec 13 '18 at 4:18




1




1




$begingroup$
First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
$endgroup$
– jgon
Dec 13 '18 at 4:19




$begingroup$
First method is wrong. Not clear how you got $a=-b$. That isn't a justified manipulation.
$endgroup$
– jgon
Dec 13 '18 at 4:19




1




1




$begingroup$
@reuns So the second one is correct?
$endgroup$
– Quoka
Dec 13 '18 at 4:22




$begingroup$
@reuns So the second one is correct?
$endgroup$
– Quoka
Dec 13 '18 at 4:22










1 Answer
1






active

oldest

votes


















3












$begingroup$

What element of $mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0rightarrow a = -b$? I'm not familiar with that element of $mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)



It might be better to think of this as row reduction, using only integer multiplication, of
$$ begin{pmatrix} 2 & 2 & 0 & 0 \ 2 & 2 & 1 & 1 end{pmatrix} $$
to get $langle 2a + 2b, 2a+2b+c+drangle = langle 2(a+b), c+d rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis
$$ (a,b,c,d) rightarrow (a,a+b,c,c+d) $$
to simplify finding your quotient).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
    $endgroup$
    – Randall
    Dec 13 '18 at 4:29






  • 1




    $begingroup$
    @Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:03










  • $begingroup$
    @MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
    $endgroup$
    – Randall
    Dec 13 '18 at 5:06








  • 1




    $begingroup$
    Yes, I see. Thank you
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:28











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1 Answer
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1 Answer
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3












$begingroup$

What element of $mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0rightarrow a = -b$? I'm not familiar with that element of $mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)



It might be better to think of this as row reduction, using only integer multiplication, of
$$ begin{pmatrix} 2 & 2 & 0 & 0 \ 2 & 2 & 1 & 1 end{pmatrix} $$
to get $langle 2a + 2b, 2a+2b+c+drangle = langle 2(a+b), c+d rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis
$$ (a,b,c,d) rightarrow (a,a+b,c,c+d) $$
to simplify finding your quotient).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
    $endgroup$
    – Randall
    Dec 13 '18 at 4:29






  • 1




    $begingroup$
    @Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:03










  • $begingroup$
    @MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
    $endgroup$
    – Randall
    Dec 13 '18 at 5:06








  • 1




    $begingroup$
    Yes, I see. Thank you
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:28
















3












$begingroup$

What element of $mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0rightarrow a = -b$? I'm not familiar with that element of $mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)



It might be better to think of this as row reduction, using only integer multiplication, of
$$ begin{pmatrix} 2 & 2 & 0 & 0 \ 2 & 2 & 1 & 1 end{pmatrix} $$
to get $langle 2a + 2b, 2a+2b+c+drangle = langle 2(a+b), c+d rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis
$$ (a,b,c,d) rightarrow (a,a+b,c,c+d) $$
to simplify finding your quotient).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
    $endgroup$
    – Randall
    Dec 13 '18 at 4:29






  • 1




    $begingroup$
    @Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:03










  • $begingroup$
    @MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
    $endgroup$
    – Randall
    Dec 13 '18 at 5:06








  • 1




    $begingroup$
    Yes, I see. Thank you
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:28














3












3








3





$begingroup$

What element of $mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0rightarrow a = -b$? I'm not familiar with that element of $mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)



It might be better to think of this as row reduction, using only integer multiplication, of
$$ begin{pmatrix} 2 & 2 & 0 & 0 \ 2 & 2 & 1 & 1 end{pmatrix} $$
to get $langle 2a + 2b, 2a+2b+c+drangle = langle 2(a+b), c+d rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis
$$ (a,b,c,d) rightarrow (a,a+b,c,c+d) $$
to simplify finding your quotient).






share|cite|improve this answer









$endgroup$



What element of $mathbb{Z}$ did you multiply by to cancel the $2$s in $2a + 2b = 0rightarrow a = -b$? I'm not familiar with that element of $mathbb{Z}$. (If I tell you that, in some group, $a^2 = b^2$, does that mean $a = b$?)



It might be better to think of this as row reduction, using only integer multiplication, of
$$ begin{pmatrix} 2 & 2 & 0 & 0 \ 2 & 2 & 1 & 1 end{pmatrix} $$
to get $langle 2a + 2b, 2a+2b+c+drangle = langle 2(a+b), c+d rangle$. Then quotient by that, as you did in your second example (in which, you made the reversible change of basis
$$ (a,b,c,d) rightarrow (a,a+b,c,c+d) $$
to simplify finding your quotient).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 4:23









Eric TowersEric Towers

32.6k22370




32.6k22370








  • 1




    $begingroup$
    I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
    $endgroup$
    – Randall
    Dec 13 '18 at 4:29






  • 1




    $begingroup$
    @Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:03










  • $begingroup$
    @MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
    $endgroup$
    – Randall
    Dec 13 '18 at 5:06








  • 1




    $begingroup$
    Yes, I see. Thank you
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:28














  • 1




    $begingroup$
    I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
    $endgroup$
    – Randall
    Dec 13 '18 at 4:29






  • 1




    $begingroup$
    @Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:03










  • $begingroup$
    @MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
    $endgroup$
    – Randall
    Dec 13 '18 at 5:06








  • 1




    $begingroup$
    Yes, I see. Thank you
    $endgroup$
    – Quoka
    Dec 13 '18 at 5:28








1




1




$begingroup$
I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
$endgroup$
– Randall
Dec 13 '18 at 4:29




$begingroup$
I don't think you can be that hard on OP for deducing $a=-b$ from $2a=-2b$. If you were in a ring theory class you'd be told that cancellation in $mathbb{Z}$ is valid even though it has few units.
$endgroup$
– Randall
Dec 13 '18 at 4:29




1




1




$begingroup$
@Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
$endgroup$
– Quoka
Dec 13 '18 at 5:03




$begingroup$
@Randall I appreciate you coming to my defense! The $a=-b$ actually came from my course solutions so I'm glad to see that my solution is the correct one
$endgroup$
– Quoka
Dec 13 '18 at 5:03












$begingroup$
@MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
$endgroup$
– Randall
Dec 13 '18 at 5:06






$begingroup$
@MathUser_NotPrime well, the other issue is that it's not an incorrect deduction. While it may be true that $a^2=b^2 Rightarrow a=b$ is false in plenty of groups, it is true in $mathbb{Z}$ (interpreted as $2a=2b$ or $a+a=b+b$).
$endgroup$
– Randall
Dec 13 '18 at 5:06






1




1




$begingroup$
Yes, I see. Thank you
$endgroup$
– Quoka
Dec 13 '18 at 5:28




$begingroup$
Yes, I see. Thank you
$endgroup$
– Quoka
Dec 13 '18 at 5:28


















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