# Thread: Quotient group question

1. ## Quotient group question

Suppose $N \unlhd G$ and $N \unlhd GN$.

I'm having trouble understanding why $GN/N \neq G/N$. Here's my reasoning:

Let $H = GN$, then $H/N = \{hN \mid h \in H \} = \{gnN \mid g \in G, n \in N \} = \{gN \mid g \in G \} = G/N$.

Where I used $nN = N$ since $n \in N$. What is wrong with this 'proof'?

2. ## Re: Quotient group question

what is the context in which this takes place (what group is GN a subset of)?

note that if N is a subgroup of G, |GN| = |G||N|/|G∩N| = |G||N|/|N| = |G|, so that in fact, GN = G.

usually, one considers two subgroups S,N normal in G. and the reason SN/N ≠ S/N is that there is no reason to suppose S/N even makes sense

(N may be "too big" to fit inside S, or may only intersect S trivially).

if one isn't considering G,N to both be subgroups of a larger group, then GN makes no sense, because it is not clear what the product should be.

on the other hand, GxN/({e}xN) is isomorphic to G, not G/N.

3. ## Re: Quotient group question

Ok, let me modify this a bit then. Let's take $H$ to be a group and let's pick a normal group, say $Z$ its center.

Say $G \leq H$, then $GZ$ is a group and $Z \unlhd GZ$. Shouldn't $GZ/Z = G/Z$ from essentially my argument above?

4. ## Re: Quotient group question

no. Z is not a subgroup of G, it's a subgroup of H. you need to find "something like Z that fits in G". that's exactly what Z∩G is.