Isomorphism of group algebras of the dihedral group.












1












$begingroup$


I'm trying to solve the following problem.




Let $m,n in mathbb{N}, m|n$. Prove that $mathbb{Z[D_{n}]}/ langle R^{m} - 1 rangle sim mathbb{Z[D_{m}]}$.




I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.



Any hint so that I can solve it will be more than helpful. Thanks a lot!






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use $langle xrangle$ for $langle xrangle$.
    $endgroup$
    – Shaun
    Jul 9 '18 at 21:35






  • 1




    $begingroup$
    Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
    $endgroup$
    – Aurel
    Jul 9 '18 at 22:13












  • $begingroup$
    @Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:28










  • $begingroup$
    @Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:50










  • $begingroup$
    So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 23:14


















1












$begingroup$


I'm trying to solve the following problem.




Let $m,n in mathbb{N}, m|n$. Prove that $mathbb{Z[D_{n}]}/ langle R^{m} - 1 rangle sim mathbb{Z[D_{m}]}$.




I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.



Any hint so that I can solve it will be more than helpful. Thanks a lot!






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use $langle xrangle$ for $langle xrangle$.
    $endgroup$
    – Shaun
    Jul 9 '18 at 21:35






  • 1




    $begingroup$
    Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
    $endgroup$
    – Aurel
    Jul 9 '18 at 22:13












  • $begingroup$
    @Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:28










  • $begingroup$
    @Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:50










  • $begingroup$
    So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 23:14
















1












1








1


1



$begingroup$


I'm trying to solve the following problem.




Let $m,n in mathbb{N}, m|n$. Prove that $mathbb{Z[D_{n}]}/ langle R^{m} - 1 rangle sim mathbb{Z[D_{m}]}$.




I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.



Any hint so that I can solve it will be more than helpful. Thanks a lot!






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




I'm trying to solve the following problem.




Let $m,n in mathbb{N}, m|n$. Prove that $mathbb{Z[D_{n}]}/ langle R^{m} - 1 rangle sim mathbb{Z[D_{m}]}$.




I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.



Any hint so that I can solve it will be more than helpful. Thanks a lot!






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 9 '18 at 21:40







Leo Lerena

















asked Jul 9 '18 at 21:34









Leo LerenaLeo Lerena

477511




477511








  • 1




    $begingroup$
    Use $langle xrangle$ for $langle xrangle$.
    $endgroup$
    – Shaun
    Jul 9 '18 at 21:35






  • 1




    $begingroup$
    Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
    $endgroup$
    – Aurel
    Jul 9 '18 at 22:13












  • $begingroup$
    @Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:28










  • $begingroup$
    @Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:50










  • $begingroup$
    So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 23:14
















  • 1




    $begingroup$
    Use $langle xrangle$ for $langle xrangle$.
    $endgroup$
    – Shaun
    Jul 9 '18 at 21:35






  • 1




    $begingroup$
    Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
    $endgroup$
    – Aurel
    Jul 9 '18 at 22:13












  • $begingroup$
    @Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:28










  • $begingroup$
    @Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 22:50










  • $begingroup$
    So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
    $endgroup$
    – Leo Lerena
    Jul 9 '18 at 23:14










1




1




$begingroup$
Use $langle xrangle$ for $langle xrangle$.
$endgroup$
– Shaun
Jul 9 '18 at 21:35




$begingroup$
Use $langle xrangle$ for $langle xrangle$.
$endgroup$
– Shaun
Jul 9 '18 at 21:35




1




1




$begingroup$
Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
$endgroup$
– Aurel
Jul 9 '18 at 22:13






$begingroup$
Hint: A map $$mathbb{Z}[mathbb{D}_n]/langle R^m -1 rangle to mathbb{Z}[mathbb{D}_m]$$ is uniquely determined by the image of the generators of $mathbb{D}_n.$ What is a reasonable choice for the image of these generators?
$endgroup$
– Aurel
Jul 9 '18 at 22:13














$begingroup$
@Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
$endgroup$
– Leo Lerena
Jul 9 '18 at 22:28




$begingroup$
@Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism.
$endgroup$
– Leo Lerena
Jul 9 '18 at 22:28












$begingroup$
@Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
$endgroup$
– Leo Lerena
Jul 9 '18 at 22:50




$begingroup$
@Aurel So I was thinking that I should send s to s and then as $langle R rangle$ is like a copy of $mathbb{Z}_n$ then it suffices to show where I send R.
$endgroup$
– Leo Lerena
Jul 9 '18 at 22:50












$begingroup$
So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
$endgroup$
– Leo Lerena
Jul 9 '18 at 23:14






$begingroup$
So that the morphism $ phi $ that sends $R^i$ to $R^j$ where $i=j mod m$. It can be clearly seen that $langle R^m -1 rangle subset Ker( phi ) $
$endgroup$
– Leo Lerena
Jul 9 '18 at 23:14












1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $m | n$ be two positive integers. Consider the morphism



$$
newcommand{zd}[1]{mathbb{Z}[mathbb{D}_{#1}]}
begin{align}
g : zd{n}& tozd{m} \
& 1 mapsto 1 \
& r mapsto rho \
& s mapsto sigma
end{align}
$$



with $r,s$ and $rho,sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $langle r^m-1 rangle subset ker g$, since



$$
g(a(r^m-1)) = g(a)(rho^m-1) = g(a)(1-1) = 0.
$$



Via the first isomorphism theorem it would suffice to see the other inclusion, so that $ker g = langle r^m -1 rangle$ and therefore $zd{n}/langle r^m-1rangle simeq zd{m}$.



Let $x = sum_{s=1}^na_sr^s + bsin zd{n}$. Now, if



$$
0 = g(x) = sum_{s=1}^na_srho^s + bsigmain zd{m},
$$



then $b = 0$ and $sum_{s=1}^na_srho^s = 0$. Noting $n = mk$, we get



$$
0 = sum_{s=1}^na_srho^s = sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}rho^{mj+i} = sum_{i=1}^mleft(sum_{j=0}^{k-1}a_{mj+i}right)rho^{i}
$$



and so



$$
sum_{j=0}^{k-1}a_{mj+i} = 0 quad (forall i). tag{$star$}
$$



Hence, we have that



$$
begin{align}
x &= sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = sum_{i=1}^mr^isum_{j=0}^{k-1}a_{mj+i}r^{mj} = sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_iright]\
& stackrel{(star)}{=} sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} -sum_{j=1}^{k-1}a_{mj+i}right] = sum_{i=1}^mr^isum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \
& = sum_{j=1}^{k-1}(r^{mj}-1)sum_{i=1}^mr^ia_{mj+i}.
end{align}
$$



To see that $x in langle r^m -1 rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) in langle r^m -1rangle$ for each $j in {1,dots,k-1}$. In effect, by the 'difference of powers' equality,



$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)sum_{l=0}^{j-1}(r^m)^l in langle r^m-1 rangle.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the detailed answer and for reviving this unanswered question!
    $endgroup$
    – Leo Lerena
    Dec 21 '18 at 1:50











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $m | n$ be two positive integers. Consider the morphism



$$
newcommand{zd}[1]{mathbb{Z}[mathbb{D}_{#1}]}
begin{align}
g : zd{n}& tozd{m} \
& 1 mapsto 1 \
& r mapsto rho \
& s mapsto sigma
end{align}
$$



with $r,s$ and $rho,sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $langle r^m-1 rangle subset ker g$, since



$$
g(a(r^m-1)) = g(a)(rho^m-1) = g(a)(1-1) = 0.
$$



Via the first isomorphism theorem it would suffice to see the other inclusion, so that $ker g = langle r^m -1 rangle$ and therefore $zd{n}/langle r^m-1rangle simeq zd{m}$.



Let $x = sum_{s=1}^na_sr^s + bsin zd{n}$. Now, if



$$
0 = g(x) = sum_{s=1}^na_srho^s + bsigmain zd{m},
$$



then $b = 0$ and $sum_{s=1}^na_srho^s = 0$. Noting $n = mk$, we get



$$
0 = sum_{s=1}^na_srho^s = sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}rho^{mj+i} = sum_{i=1}^mleft(sum_{j=0}^{k-1}a_{mj+i}right)rho^{i}
$$



and so



$$
sum_{j=0}^{k-1}a_{mj+i} = 0 quad (forall i). tag{$star$}
$$



Hence, we have that



$$
begin{align}
x &= sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = sum_{i=1}^mr^isum_{j=0}^{k-1}a_{mj+i}r^{mj} = sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_iright]\
& stackrel{(star)}{=} sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} -sum_{j=1}^{k-1}a_{mj+i}right] = sum_{i=1}^mr^isum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \
& = sum_{j=1}^{k-1}(r^{mj}-1)sum_{i=1}^mr^ia_{mj+i}.
end{align}
$$



To see that $x in langle r^m -1 rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) in langle r^m -1rangle$ for each $j in {1,dots,k-1}$. In effect, by the 'difference of powers' equality,



$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)sum_{l=0}^{j-1}(r^m)^l in langle r^m-1 rangle.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the detailed answer and for reviving this unanswered question!
    $endgroup$
    – Leo Lerena
    Dec 21 '18 at 1:50
















1












$begingroup$

Let $m | n$ be two positive integers. Consider the morphism



$$
newcommand{zd}[1]{mathbb{Z}[mathbb{D}_{#1}]}
begin{align}
g : zd{n}& tozd{m} \
& 1 mapsto 1 \
& r mapsto rho \
& s mapsto sigma
end{align}
$$



with $r,s$ and $rho,sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $langle r^m-1 rangle subset ker g$, since



$$
g(a(r^m-1)) = g(a)(rho^m-1) = g(a)(1-1) = 0.
$$



Via the first isomorphism theorem it would suffice to see the other inclusion, so that $ker g = langle r^m -1 rangle$ and therefore $zd{n}/langle r^m-1rangle simeq zd{m}$.



Let $x = sum_{s=1}^na_sr^s + bsin zd{n}$. Now, if



$$
0 = g(x) = sum_{s=1}^na_srho^s + bsigmain zd{m},
$$



then $b = 0$ and $sum_{s=1}^na_srho^s = 0$. Noting $n = mk$, we get



$$
0 = sum_{s=1}^na_srho^s = sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}rho^{mj+i} = sum_{i=1}^mleft(sum_{j=0}^{k-1}a_{mj+i}right)rho^{i}
$$



and so



$$
sum_{j=0}^{k-1}a_{mj+i} = 0 quad (forall i). tag{$star$}
$$



Hence, we have that



$$
begin{align}
x &= sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = sum_{i=1}^mr^isum_{j=0}^{k-1}a_{mj+i}r^{mj} = sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_iright]\
& stackrel{(star)}{=} sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} -sum_{j=1}^{k-1}a_{mj+i}right] = sum_{i=1}^mr^isum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \
& = sum_{j=1}^{k-1}(r^{mj}-1)sum_{i=1}^mr^ia_{mj+i}.
end{align}
$$



To see that $x in langle r^m -1 rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) in langle r^m -1rangle$ for each $j in {1,dots,k-1}$. In effect, by the 'difference of powers' equality,



$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)sum_{l=0}^{j-1}(r^m)^l in langle r^m-1 rangle.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the detailed answer and for reviving this unanswered question!
    $endgroup$
    – Leo Lerena
    Dec 21 '18 at 1:50














1












1








1





$begingroup$

Let $m | n$ be two positive integers. Consider the morphism



$$
newcommand{zd}[1]{mathbb{Z}[mathbb{D}_{#1}]}
begin{align}
g : zd{n}& tozd{m} \
& 1 mapsto 1 \
& r mapsto rho \
& s mapsto sigma
end{align}
$$



with $r,s$ and $rho,sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $langle r^m-1 rangle subset ker g$, since



$$
g(a(r^m-1)) = g(a)(rho^m-1) = g(a)(1-1) = 0.
$$



Via the first isomorphism theorem it would suffice to see the other inclusion, so that $ker g = langle r^m -1 rangle$ and therefore $zd{n}/langle r^m-1rangle simeq zd{m}$.



Let $x = sum_{s=1}^na_sr^s + bsin zd{n}$. Now, if



$$
0 = g(x) = sum_{s=1}^na_srho^s + bsigmain zd{m},
$$



then $b = 0$ and $sum_{s=1}^na_srho^s = 0$. Noting $n = mk$, we get



$$
0 = sum_{s=1}^na_srho^s = sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}rho^{mj+i} = sum_{i=1}^mleft(sum_{j=0}^{k-1}a_{mj+i}right)rho^{i}
$$



and so



$$
sum_{j=0}^{k-1}a_{mj+i} = 0 quad (forall i). tag{$star$}
$$



Hence, we have that



$$
begin{align}
x &= sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = sum_{i=1}^mr^isum_{j=0}^{k-1}a_{mj+i}r^{mj} = sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_iright]\
& stackrel{(star)}{=} sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} -sum_{j=1}^{k-1}a_{mj+i}right] = sum_{i=1}^mr^isum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \
& = sum_{j=1}^{k-1}(r^{mj}-1)sum_{i=1}^mr^ia_{mj+i}.
end{align}
$$



To see that $x in langle r^m -1 rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) in langle r^m -1rangle$ for each $j in {1,dots,k-1}$. In effect, by the 'difference of powers' equality,



$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)sum_{l=0}^{j-1}(r^m)^l in langle r^m-1 rangle.
$$






share|cite|improve this answer











$endgroup$



Let $m | n$ be two positive integers. Consider the morphism



$$
newcommand{zd}[1]{mathbb{Z}[mathbb{D}_{#1}]}
begin{align}
g : zd{n}& tozd{m} \
& 1 mapsto 1 \
& r mapsto rho \
& s mapsto sigma
end{align}
$$



with $r,s$ and $rho,sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $langle r^m-1 rangle subset ker g$, since



$$
g(a(r^m-1)) = g(a)(rho^m-1) = g(a)(1-1) = 0.
$$



Via the first isomorphism theorem it would suffice to see the other inclusion, so that $ker g = langle r^m -1 rangle$ and therefore $zd{n}/langle r^m-1rangle simeq zd{m}$.



Let $x = sum_{s=1}^na_sr^s + bsin zd{n}$. Now, if



$$
0 = g(x) = sum_{s=1}^na_srho^s + bsigmain zd{m},
$$



then $b = 0$ and $sum_{s=1}^na_srho^s = 0$. Noting $n = mk$, we get



$$
0 = sum_{s=1}^na_srho^s = sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}rho^{mj+i} = sum_{i=1}^mleft(sum_{j=0}^{k-1}a_{mj+i}right)rho^{i}
$$



and so



$$
sum_{j=0}^{k-1}a_{mj+i} = 0 quad (forall i). tag{$star$}
$$



Hence, we have that



$$
begin{align}
x &= sum_{i=1}^msum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = sum_{i=1}^mr^isum_{j=0}^{k-1}a_{mj+i}r^{mj} = sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_iright]\
& stackrel{(star)}{=} sum_{i=1}^mr^ileft[sum_{j=1}^{k-1}a_{mj+i}r^{mj} -sum_{j=1}^{k-1}a_{mj+i}right] = sum_{i=1}^mr^isum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \
& = sum_{j=1}^{k-1}(r^{mj}-1)sum_{i=1}^mr^ia_{mj+i}.
end{align}
$$



To see that $x in langle r^m -1 rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) in langle r^m -1rangle$ for each $j in {1,dots,k-1}$. In effect, by the 'difference of powers' equality,



$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)sum_{l=0}^{j-1}(r^m)^l in langle r^m-1 rangle.
$$







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edited Dec 20 '18 at 11:25

























answered Dec 20 '18 at 11:19









Guido A.Guido A.

8,0701730




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  • $begingroup$
    Thanks for the detailed answer and for reviving this unanswered question!
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    – Leo Lerena
    Dec 21 '18 at 1:50


















  • $begingroup$
    Thanks for the detailed answer and for reviving this unanswered question!
    $endgroup$
    – Leo Lerena
    Dec 21 '18 at 1:50
















$begingroup$
Thanks for the detailed answer and for reviving this unanswered question!
$endgroup$
– Leo Lerena
Dec 21 '18 at 1:50




$begingroup$
Thanks for the detailed answer and for reviving this unanswered question!
$endgroup$
– Leo Lerena
Dec 21 '18 at 1:50


















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