TITLE: $A/B \cong A \iff B = \{e\}$ for groups $A,B$?
QUESTION [6 upvotes]: 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

REPLY [3 votes]: This is not necessarily the case.
Counterexample: take
$$
A = \bigoplus_{i \in \Bbb N} \Bbb Z_2, \quad
B = \Bbb Z_2
$$

REPLY [3 votes]: 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 $\mathbb Z$.

REPLY [2 votes]: This is not true, consider the group $\mathbb Q/\mathbb Z$. Now consider the subgroup $\{0,\frac{1}{2}\}$.
We have $\mathbb Q/\mathbb Z\cong (\mathbb Q/\mathbb Z)/\{0,\frac{1}{2}\}$