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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.

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Does |G/H| = |G| / |H| help? –  Jack Schmidt Oct 19 '11 at 17:06
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I'm not entirely sure what you are asking...however, working out $|M/N|$ isn't too hard. One can think of this (sloppily) as $|M|/|N|$. So if $M$ is finite then so are $N$ and $M/N$. Similarly, if $M$ is an infinite group but $N$ is finite then $M/N$ is infinite. If both of $M$ and $N$ are infinite then $M/N$ can be either finite or infinite. The fact that both can occur is a not-too-hard exercise... –  user1729 Oct 19 '11 at 17:11
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@iyengar: As far as I know, your formula for Sha is not correct in general; Theorem X.4.2 of Silverman's "Arithmetic of Elliptic Curves" says that if $\phi\colon E/K\to E'/K$ is an isogeny of elliptic curves, then you have an exact sequence $$0\to E'(K)/\phi(E(K)) \to S^{(\phi)}(E/K) \to \text{Ш}(E/K)[\phi]\to 0,$$and you seem to want to take $\phi$ to be multiplication by $2$. But the group on the right is not all of the Tate-Shafarevich group, it's only the group you obtain by restricting to the kernel of the isogeny, and the middle part is not the whole Selmer group either. –  Arturo Magidin Oct 19 '11 at 17:23
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@iyengar: I don't know enough to comment: I can only point you to the fact that your claim seems to disagree with the theorem I quoted, to note that you yourself say that nobody seems to be able to confirm your formula, and to echo Mariano in saying "if the expression were so simple, and the result so stupid, somebody would have noticed by now." And forgive me if I find it somewhat incredible that you can find a simple expression for the group Sha when you (i) confess you don't really know what Sha is, and (ii) you don't know the rather basic fact that a quotient of a finite group is finite. –  Arturo Magidin Oct 19 '11 at 17:29
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I did not downvote anything here, but I really wish you spent an extra 2 minutes making sure what you write is at least properly spaced---it is not hard to follow the very trivial rules of capitalization and spacing between words and punctuation, and the improvement in readability would be immense (and there is rarely any need to write more than one question mark!) –  Mariano Suárez-Alvarez Oct 19 '11 at 17:35

1 Answer 1

up vote 3 down vote accepted

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?

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Then it will imply the finiteness of Tate-Shafarevich Group sir –  Iyengar Oct 19 '11 at 17:19
3  
@iyengar: Only if your claimed formula is correct in general, which does not seem to me to be the case. Check Theorem X.4.2., p 298, of Silverman's "Arithmetic of Elliptic Curves". The quotient you give is only the $[2]$ part of the Shafarevich group, not the entire thing, and you don't have the entire Selmer group, you only have the $\phi$ part. You yourself note that you "fear the expression may not be true." I would wager it is not true. –  Arturo Magidin Oct 19 '11 at 17:25
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@iyengar: don't you think that if the conjecture would be that stupid, someone would have notices a few years ago? ... –  Mariano Suárez-Alvarez Oct 19 '11 at 17:25
    
@MarianoSuárez-Alvarez:ya thats what i mentioned in my question,read it again,i also said that it wont turn to be so easy in my last edit,if it does turn,then answer me why it wont turn ,ok??? –  Iyengar Oct 19 '11 at 17:30
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@iyengar: I have to say, your entire "connection to Sha" reads to me like someone saying that they think they have a proof 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. –  Arturo Magidin Oct 20 '11 at 0:14

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