Confusion about the difference between HN/N and H/N in the second isomorphism theorem

Question

Im trying to understand the second isomorphism theorem but I am stuck. So in my textbook the author did the following:

Let G be a group and N<G a normal subgroup and H<G a subgroup. Let $\pi: G\rightarrow G/N$ be the canonical homomorphism given by $g\mapsto gN$. If we limit this map to H, then we have homomorphism $H\mapsto G/N$ with kernel $H\cap N $ and image HN/N. I understand why the kernel is equal to $H\cap N$, but why is the image equal to HN/N?. What I think is the following: The homomorphism $H\rightarrow G/N$ is defined as $h\mapsto hN$ which means the image for the whole subgroup H is $H/N=\{hN|h\in H\}$. Now the problem is that N can not be a subgroup of H so H/N does not make sense. Is that the reason why we have HN/N instead, because they are the same set $HN/N=\{hnN|h\in H\}=\{hN|h\in H\}=H/N$ or are they not? If not what is the difference between them?

Answer 1

You are of course correct that the image of ${\left.\pi\right|_H}$ is ${\{hN: h \in H\}}$, which is a subgroup of ${G/N}$ with group structure given by ${(hN)(h'N) = (hh')N}$. You could refer to this as ${H/N}$ if you like, but we tend to strictly reserve the notation of "${A/B}$" when $B$ is a sub-structure of $A$, which, as you point out, is not necessarily the case in this context.

If instead you consider the product group ${HN}$, then $N$ really is then a normal subgroup of ${HN}$, and so the notation ${(HN)/N}$ follows the standard convention mentioned above. Furthermore, ${(HN)/N}$ as a group is clearly the same as ${\{hN: h \in H\}}$ with the above mentioned product.

To summarise: we use ${(HN)/N}$ instead of ${H/N}$ to keep up with standard notation conventions.

Answer 2

Consider an actual example: $\pi : \mathbf Z \to \mathbf Z/24\mathbf Z$ where $G = \mathbf Z$ and $N = 24\mathbf Z$. Set $H = 10\mathbf Z$. Here we're using additive groups, so the notation becomes additive ($H + N$ instead of $HN$). The image of $H$ is what you get by repeatedly adding $10$ to itself modulo $24$: \begin{align*} \{10, 20, 6, 16, 2, 12, 22, 8, 18, 4, 14, 0\} \bmod 24 & = \{0, 2, 4, 6, 8, 10, 12, 14, 16, 18\} \bmod 24 \\ & = 2\mathbf Z/24\mathbf Z, \end{align*} where $2\mathbf Z = 10\mathbf Z + 24\mathbf Z$ (since $2 = \gcd(10,24)$) $= H + N$, so the image of $\pi$ is $(H+N)/N$. Would you call the image $H/N = 10\mathbf Z/24\mathbf Z$?