Question regarding the image of the homomorphism in the 2nd isomorphism theorem

Question

Let $G$ be a group, $H$ a subgroup, $N$ a normal subgroup of $G$. Let $f$ be the map $$f: H \rightarrow G/N, \: h \mapsto hN$$ I am trying to find the image of this homomorphism:

$Im(f) = \{hN \: | h \in H\} = H/N.\:$ However, in my textbook the image is written as $HN/N. \:$ I understood the two are equivalent since $$HN/N = \{hnN \: | h \in H, n \in N\} = \{hN \: | h \in H, n \in N\} = \{hN \: | h \in H\} = H/N$$

but I'm wondering why $HN/N$ is written instead of $H/N$.

Answer 1

Let’s view this from a mere set-theoretic level first. Stricty speaking, one only defines for a group $G$ and a subgroup $N ⊆ G$ $$G/N = \{gN;~g ∈ G\}.$$ Of course, it would be also okay to further define for subgroups $H ⊆ G$ $$H/N = \{hN;~h ∈ H\}.$$ This seems analogous. However, this kind of subverts the two real-life interpretations of the notation “$G/N$”.

1. Group actions. More generally, if a group $N$ is acting on a set $X$ from the right, then the orbit space of this action is denoted by $$X/N = \{xN;~x ∈ X\}.$$ (For a left action of $N$ on $X$, thee orbit space is $N\backslash X = \{Nx;~x ∈ X\}$.)

2. Factor groups. To denote the factor group of a group $G$ by a normal subgroup $N$, we write $$G/N = \{[g]_N;~g ∈ N\}$$ and mean by it a set endowed with a group structure.

For 1.), we need $N$ to act on $H$ from the right, but the right action from $N$ on $G$ doesn’t necessarily restrict to $H$. There might be $h ∈ H$ and $n ∈ N$ with $hn \notin H$. In fact, when $N \not ⊆ H$, there is some $n ∈ N$ with $n \notin H$, so for $h = 1$, $hn \notin H$. Hence, we need $H ⊆ N$ here.

For 2), we need $N$ to be a subgroup of $H$ by definition. So again we need $H ⊆ N$ here.

Hence, $H/N$ neither is an orbit space nor a factor group in general, but only if $H ⊆ N$.

Also note that the orbit space $G/N$ of the right action of $N$ on $G$ gives an implementation of the factor group $G/N$, which is why we use the same notation for both concepts. They fit together seamlessly.

So, all in all, the notation only makes sense in general when $H ⊆ N$, which is why one chooses to enlarge the group $H$ to $HN$ (which really is a subgroup since $N$ is normal) and write “$HN/N$” instead of “$H/N$”. As you already checked, this set gives the correct image.