$A/B \cong A \iff B = \{e\}$ for groups $A,B$? Let $A$ and $B$ be groups (can be infinite)
Is it true that $A/B$ isomorphic to $A$ $\Leftrightarrow$ $B=\{e\}$ 
I didn't find a way to prove it.
Thanks
 A: Not quite true as stated. What is true is that the canonical map $A\to A/B$ is isomorphism if and only if $B=\{e\}$. On the other hand, take for $A$ the circle group, which you can think of as all complex numbers of absolute value $1$. Now choose $n>1$ and map $z\to z^n$; then the kernel $B$ will be the group of $n$-th roots of unity, but $A/B\simeq A$.
A: This is not necessarily the case.
Counterexample: take
$$
A = \bigoplus_{i \in \Bbb N} \Bbb Z_2, \quad
B = \Bbb Z_2
$$
A: If $B = \{e\}$ then certainly $A/B \simeq A$.  The isomorphism sends the coset $a + \{e\}$ to the element $a$.
Conversely if the natural map $A \to A/B$ is an isomorphism then $B = \{e\}$ because $B$ is the kernel of this map.  More generally if $A$ is finite and $A/B \simeq A$ then $B = \{e\}$ because if $|A/B| = |A|$ then $|B| = 1$.  But if $A$ is infinite then this need not be true.  For example take a countable product $A = \mathbb{Z \times Z \times \cdots}$ and quotient by the first factor $B = \mathbb Z \times 0 \times 0 \times \cdots$, the result is still a countable product of $\mathbb Z$.
A: This is not true, consider the group $\mathbb Q/\mathbb Z$. Now consider the subgroup $\{0,\frac{1}{2}\}$.
We have $\mathbb Q/\mathbb Z\cong (\mathbb Q/\mathbb Z)/\{0,\frac{1}{2}\}$
A: This is not true: In fact, as abelian groups, $\mathbb R/\mathbb Q\cong \mathbb R$. This is because both are vector spaces over $\mathbb Q$, and both have uncountable cardinality, and thus must be of dimension $\mathfrak c$. Thus, they are isomorphic as vector spaces and in particular, as abelian groups.
