1
$\begingroup$

This question is an interesting one,not like my previous one.

Can we judge the size of a Quotient Group by seeing the size of its constituents ?

To add something ,Suppose consider a group $\rm{G}$ which is defined as $\rm{G=M/N}$ (Quotient Group),then my Question is can we comment the Cardinality(whether finite or not) of the group $\rm{G}$ by knowing that $\rm{M,N}$ are finite.

This Question has a good implication.

This question was actually a result of my analysis of Tate-shafarevich group. We know by the Mordell-Weil Theorem that $E(\mathbb{Q})/\rm{2}E(\mathbb{Q})$ is finite(which is phrased as Weak Mordell-Weil Theorem).

And also we know that "Selmer Group which is defined as the principal homogeneous spaces that $K_v$-rational points for all places $v$ of $K$ is finite .

But I understood that

$Ш(E/\mathbb{Q})=[\rm{Sel}(E/\mathbb{Q})/(E(\mathbb{Q})/\rm{2}E(\mathbb{Q})]$ * And the Finiteness of Tate-Shafarevich Group is an Open question , So we know that as Selmer Group and $E(\mathbb{Q})/\rm{2}E(\mathbb{Q})$ are finite ,So can we comment about the finiteness of $Ш(E/\mathbb{Q})$?

Like some one told that
If $G=M/N$ and $M$ is finite, then $G$ must be finite: quotient of a finite group is necessarily finite.

But according to that ,it will imply that the Group $Ш(E/\mathbb{Q})$ will be finite,as Selmer and $E(\mathbb{Q})/\rm{2}E(\mathbb{Q})$ are finite,which solves the problem in determining the finiteness of Tate-Shafarevich Group.But i think that this is not the case,as its an Age-old problem,and it doesnt turn out to be so simple


*:That expression of Tate-Shafarevich Group was a result of my thinking ,and i fear that it may not be true,as no one neither in Stack exchange nor in Mathoverflow helped me in knowing what is $Ш(E/\mathbb{Q})$.


Note:Anyone putting downvotes are humbly requested to post the reason,which helps in fixing my errors and moulding me,

Thanks everyone, Cordially, Iyengar.

  • 0
    Now I realized that i was talking about 2-selmer group,like you have said2011-10-20

1 Answers 1

3

For any group $G$, if $M$ is a group, $N$ is a subgroup of $M$, and $G\cong M/N$, then $|G||N| = |M|$ in the sense of cardinality.

This because the underlying set of $G$ is the set of equivalence classes of $M$ modulo $N$. These equivalence classes partition $M$, so if $\{m_i\}_{i\in I}$ are a complete set of coset representatives for $N$ in $M$, then $|M| = \left|\bigcup_{i\in I}m_iN\right| = \sum_{i\in I}|m_iN| = \sum_{i\in I}|N| = |I||N|,$ the next-to-last equality because $mN$ is bijectable with $N$ for every $m\in M$. Since the number of equivalence classes is $I$, $|G|=|I|$, so we get $|G||N|=|M|$.

In particular, if $M$ is finite, then necessarily $G$ is finite.

Are you sure this is really what you wanted to ask?

  • 6
    @iyengar: I have to say, your entire "connection to Sha" reads to me like someone saying that they think they have a $p$roof of the Riemann hypothesis, mentioning along the way that they aren't actually very sure about what "the Riemann zeta function" is, and saying that their proof just hinges on whether one can or cannot say that the product of two real numbers is always a real number, or if it sometimes can be a complex number with nonzero imaginary part. In short, not something that can be taken seriously.2011-10-20