I have a group $G$ and two normal subgroups $N_1, N_2 \unlhd G$ with $N_1 \leq N_2$.

Is it true that $G/N_2 \leq G/N_1$?

  • $\begingroup$ If $\leq$ means subgroup then it is true. $\endgroup$ – Jihad Jan 4 '15 at 8:24
  • $\begingroup$ What do you mean by $N_1 \le N_2$? $\endgroup$ – Frog Jan 4 '15 at 8:25
  • 2
    $\begingroup$ Hint: See what happens when one of the subgroups is trivial. $\endgroup$ – Tobias Kildetoft Jan 4 '15 at 8:28
  • $\begingroup$ @TobiasKildetoft I don't get this hint $\endgroup$ – user204667 Jan 4 '15 at 8:35
  • $\begingroup$ Then you seem to mean something different with $\leq $ than the usual meaning of "subgroup of". $\endgroup$ – Tobias Kildetoft Jan 4 '15 at 8:36

Along the lines of what is mentioned in the comments, consider $G=\mathbb Z$, $N_1=\{0\}$, $N_2=2\mathbb Z$. $G/N_1\cong \mathbb Z$ and $G/N_2\cong\mathbb Z_2$, and neither of these groups is isomorphic to a subgroup of the other.


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