Do alternative expressions exist for subgroups of $G/N$, distinct from the correspondence theorem for groups?

Question

If $G$ is a group and $N$ is a normal subgroup, then any subgroup $J$ of $G/N$ can be expressed as $J=H/N$ for some subgroup $H$ of $G$ containing $N$ ($N \leq H \leq G$). For proof, see Subgroup of factor/quotient group $G/N$ is of form $H/N$ for $N \subseteq H$. Furthermore, there is a bijetion between subgroups of $G/N$ and subgroups of $G$ containing $N$.

What confuses me is this: might we be able to find a group $W$ not containing $N$ s.t. $W/N \leq G/N$? Of course, we still must have $W/N = H/N$ for some group $H$ satisfying $N \leq H \leq G$ by the correspondence theorem above.

In other words, for subgroup $H/N \leq G/N$, is it possible that $H / N = W/N$, where group $W \neq H$ does not contain $N$?

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EDIT: a factor/quotient group $G/N$ is defined when $N$ is a normal subgroup of $G$. So if $N \nsubseteq W$, then $W/N$ does not make sense as a quotient group. The question should be rephrased in terms of sets: is it possible that $H/N=\{hN \mid h \in H \leq G\} = \{wN \mid w \in W \leq G \}$? For example, $nN=N=eN$, so $W$ need not include $n$ if it already contains $e$. Based on that last sentence, it is certainly possible to have $H/N = \{wN \mid w \in W \subseteq G\}$, but $W$ may not be a group.

Answer 1

You have fixed $G$, a normal subgroup $N$, and the map $f: G\to G/N$ such that $f(x)=xN$. So you're asking about the equality $f(W)=f(H)$ where $H$ is a subgroup containing $N$, and $W$ is possibly something other than $H$.

You've already noted that this is possible if $W$ is allowed to be a subset. But we can do it with a subgroup too. First maybe we should also assume that $N$ is a proper subgroup of $H$, since otherwise it's a little boring as we can take $W=\{e\}$. So let's assume this from now on.

Observation 1. If $f(W)=f(H)$ then $W\subseteq H$.

Proof. Fix $w\in W$. Then $wN=hN$ for some $h\in H$. So $h^{-1}w\in N$, which means it's in $H$. So $w\in H$.

Observation 2. If $W\subseteq H$ then $f(W)=f(H)$ if and only if $W\cap hN\neq\emptyset$ for all $h\in H$.

Proof. If $f(W)=f(H)$ then for any $h\in H$ there is $w\in W$ such that $wN=hN$ and so $w\in hN$. For the other direction, suppose $W\cap hN\neq\emptyset$ for all $h\in H$. $f(W)\subseteq f(H)$ is obvious. Given $h\in H$, choose $w\in W\cap hN$. Then $wN=hN$ so $f(h)\in f(W)$.

So now we see that to make a counterexample with a subgroup we just need a subgroup that hits all of the $H$-cosets of $N$ but doesn't contain everything in $N$.

Example 3. Probably the simplest example is in $G=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Let $H=G$ and $N=\{(0,0),(1,0)\}$. Then take $W=\{(0,0),(0,1)\}$.

Observation 4. If $W\subseteq N$ then $f(W)=f(H)$ if and only if $H=WN$.

Proof. This is really just a different way to write Observation 2.

So we can get examples whenever we have subgroups $W$ and $N$ of $H$ such that $H=WN$ but $W$ does not contain $N$. This happens a lot in finite abelian groups since the whole structure theorem is based off of writing an abelian group $H$ as $H=WN$ where $W\cap N=\{e\}$. For example, if $H=\mathbb{Z}_{m}\times\mathbb{Z}_n$ then take $N=\mathbb{Z}_m\times\{0\}$ and $W=\{0\}\times\mathbb{Z}_n$.

Another example is $G=H=\mathbb{Z}$, $N=3\mathbb{Z}$ and $W=5\mathbb{Z}$. In general take $N=m\mathbb{Z}$ and $W=n\mathbb{Z}$, with $m,n>1$ and relatively prime.

So is it always possible for a given $G$, $H$, and $N$ to find a such a $W$? Well, no. One way to ruin it is by taking $H$ to have the property that its subgroups are linearly ordered by inclusion. In this case, if $W$ does not contain $N$ then $N$ contains $W$, so if $f(H)=f(W)$ then $H=WN=N$. Groups with this property include finite cyclic groups of prime power order and Prufer p-groups.

Answer 2

If $G$ is a group, $N$ is a normal subgroup of $G$, then you have two families of subgroups of $G$, and both are connected to subgroups of $G/N$ via the isomorphism theorems:

1. The subgroups $H$ of $G$ with $N\subseteq H$. The Fourth/Lattice Isomorphism Theorem tells you that these subgroups are in one-to-one, inclusion preserving correspondence with the subgroups of $G/N$.

2. The subgroups $H$ of $G$ that do not contain $N$. Here you use the Second Isomorphism Theorem: we know that restricting the map $G\to G/N$ to $H$ will give you a homomorphism with kernel $H\cap N$. And the Second Isomorphism Theorem tells you that its image, which is isomorphic to $H/(N\cap H)$ by the First Isomorphism Theorem, satisfies $$\frac{H}{N\cap H} \cong \frac{HN}{N}.$$ Note that $HN$ is a subgroup of $G$ that contains $N$, and so is of the first kind.

So the isomorphism theorems tell you exactly what happens to every subgroup of $G$ under the map $G\longmapsto G/N$: those that $H$ contain $N$ are mapped to the subgroups $H/N$ of $G/N$ in the “natural way”; and those $H$ that do not contain $N$ are mapped to subgroups isomorphic their own quotients $H/(N\cap H)$, and to the same thing as $HN$, that is, to $HN/N$. And if you trace the proof of the Second Isomorphism Theorem, you will see that the map is again the “obvious one”, $h\longmapsto hN$.