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.
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?