Given a group $G$ and its normal subgroups $N_1$ and $N_2$, s.t. $N_1\subset N_2$. Prove that there exists a subgroup $H$ of $G/N_1$, such that $ H \simeq G/N_2$.
There is a one Claiming the above statement is right.
In my thought, this statement looks like a false.
But Can't find Any counterexample of the above statement.
Then, Is that true statement?
As a hint: $N_1$ is a red herring. If $(G,N_1,N_2)$ is a counterexample, then so is $(G/N_1,1,N_2/N_1)$, so you may take $N_1=1$.
Using the classification of finitely generated abelian groups, you can prove the statement is true for abelian groups, so a good start is to look at small non-abelian groups.
The smallest counterexample (I think) has $|G|=8$, $N_1=1$, $|N_2|=2$