Let N1 and N2 are normal subgroups in the finite group G. Is it true that if N1≃N1 then G∖N1≃G∖N2.?

Question

Let $N_1$ and $N_2$ are normal subgroups in the finite group $G$. Is it true that if $N_1 \simeq N_2$ then $G/ N_1 \simeq G/ N_2$?

Answer 1

No.

Think of $G = \langle a \rangle \times \langle b \rangle$, where $a$ has order $4$, and $b$ has order $2$.

The quotient of $G$ modulo $\langle b \rangle$ is cyclic of order $4$, while in the quotient modulo $\langle a^2 \rangle$ all non-identity elements have order $2$.

Still $\langle a^2 \rangle \cong \langle b \rangle$, as they are both cyclic of order $2$.

Answer 2

No, take $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, with $H=\langle (1,0)\rangle$ and $K=\langle (0,2)\rangle$

Answer 3

However you wanted $G$ to be finite and @Tobias remaked me, you can keep my small hint when the group is infinite. $$N_1=2\mathbb Z$$ and $$N_2=3\mathbb Z$$ while $G=\mathbb Z$.