I'm studying linear Algebra by myself right now and I'm currently reading about quotient groups. I wanted to disprove the following: for two normal subgroups $N_1\subseteq N_2$ of $(G,\cdot)$, $G/N_2$ is a subgroup of $G/N_1$.
I wanted to disprove it by giving a counterexample and thought of $N_1=\{e\}\subseteq G=N_2$ and $G/N_1=G/\{e\}=G$ and $G/N_2=G/G=\{G\}$. Since $\{G\}\not\subseteq G$ it is not a subgroup.
Then it occurred to me, shouldn't it technically be $G/\{e\}=\{\{g\} | g\in G\}$, because we are looking at the cosets of $\{e\}$, which each contain one element but still are sets and not just the element itself? It doesn't really change the proof I'm just confused.
Also, when an exercise asks whether one group is a subgroup like in this example, is it (if not specified) really talking about real subgroups or more like isomorphic to a subgroup, generally speaking.
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2$\begingroup$ I think they want you to show an example where $G/N_2$ does not embed as a subgroup of $G/N_1$. If $N_2$ is a strict superset of $N_1$ then $G/N_2$ is not a subset of $G/N_1$ - so it's not a subgroup either. But this is too trivial. I think the exercise means something less trivial. $\endgroup$Chad K– Chad K2025-11-21 16:30:25 +00:00Commented 18 hours ago
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Good question. Formally speaking $G/N_2$ is a subgroup of $G/N_1$ only when $N_1 = N_2$, since otherwise the elements of $G/N_2$ are cosets of $N_2$, which are not elements of $G/N_1$ (which are cosets of $N_1$).
But (like you) I suspect that they are looking for an example where $G/N_2$ is not isomorphic to a subgroup of $G/N_1$, and if so they should certainly have made that clear.
It's easiest to look for an example with $N_1 = \{e\}$, and a standard example is $G = Q_8$ (the quaternion group) and $N_2$ the (unique) subgroup of order $2$.
