Suppose $AN_1=AN_2=G$, $A\cap N_1=A\cap N_2=1$ and $N_1,N_2$ are normal in $G$, do we have $N_1\cong N_2$?

Let $$G$$ be a finite group. Suppose $$AN_1=AN_2=G$$, $$A$$ is a subgroup of $$G$$, $$A\cap N_1=A\cap N_2=1$$ and $$N_1,N_2$$ are normal in $$G$$, do we have $$N_1\cong N_2$$?

I think this is not true,but I failed to find an counterexample.

I know $$G/N_1\cong G/N_2$$ can't imply $$N_1\cong N_2$$, for example, $$G=D_8, N_1=C_4,N_2=C_2\times C_2.$$

• What does $A$ stand for? – Arthur Jun 20 at 8:45
• Sorry I forgot to mention. It's a subgroup of $G$. – abvdd Jun 20 at 8:51

As you did, $$G = D_8 = \langle r,s:r^4 = s^2 = (rs)^2 = 1\rangle$$, $$N_1 = \langle r\rangle\cong C_4$$ and $$N_2 = \langle r^2,s\rangle\cong C_2\times C_2$$. Both $$N_1$$ and $$N_2$$ are normal in $$G$$ because they both have index $$2$$. Now define $$A = \langle rs\rangle\cong C_2$$. We have $$AN_1 = G = AN_2$$ and $$A\cap N_1 = 1 = A\cap N_2$$ but $$N_1$$ and $$N_2$$ are not isomorphic.