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, $G=D_8, N_1=C_4,N_2=C_2\times C_2.$

  • $\begingroup$ What does $A$ stand for? $\endgroup$
    – Arthur
    Jun 20, 2019 at 8:45
  • $\begingroup$ Sorry I forgot to mention. It's a subgroup of $G$. $\endgroup$
    – abvdd
    Jun 20, 2019 at 8:51

1 Answer 1


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.


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