# Proof of 2nd and 3rd isomorphism theorems

I am doing an exercise in Artin's "Algebra" that asks to prove the 2nd and 3rd isomorphism theorems. They don't seem that complicated, I'm pretty sure that what I'm doing is correct, I just wanted to validate that and iron out some kinks.

To prove the 2nd theorem (given a subgroup $$H$$ and normal subgroup $$N$$ $$\Rightarrow$$ $$H/H\cap N\cong HN/N$$), restrict the canonical homomorphism $$\pi:G\to G/N$$ to $$H$$. Then the kernel will be $$H\cap N$$ (because $$N$$ is the kernel of $$\pi$$). Then, by the first isomorphism theorem, $$H/H\cap N\cong im (\pi_H)$$. I am just not quite sure why the image has to be $$HN/N$$. Is it because every $$h$$ in $$H$$ is taken to a coset $$hN$$ and $$HN/N$$ is the set of those cosets? Why not $$H/N$$ then?

To prove the 3rd theorem (given normal subgroups $$H$$ and $$N$$ s.t. $$H\supset N$$ $$\Rightarrow$$ $$G/H\cong \bar G /\bar H$$, where $$\bar G=G/N$$ and $$\bar H=H/N$$), first show that $$\bar H$$ is normal in $$\bar G$$. I thought this could be done through the correspondence theorem (given a surjective homomorphism between 2 groups, there is a correspondence between subgroups of the 1st that contain the kernel and the subgroups of the 2nd, and the normality of one implies the normality of the other). We have this by the canonical surjective map $$\pi:G\to \bar G$$, and since $$H$$ contains the kernel ($$N$$) and is normal, its corresponding subgroup (the image of $$H$$ under $$\pi$$) $$\bar H$$ is also normal. Since $$\bar H$$ is normal $$\bar G/\bar H$$ makes sense and we also have a canonical surjective map $$\rho:\bar G\to \bar G/\bar H$$. Therefore the composition $$\rho\circ \pi:G\to \bar G/\bar H$$ is also a surjective homomorphism and $$G/ker(\rho\circ \pi)\cong \bar G/\bar H$$ and the kernel of $$\rho\circ \pi$$ is $$H$$ ($$\bar H$$ is the kernel of $$\rho$$ and $$\pi^{-1}(\bar H)=H$$). Therefore $$G/H\cong \bar G /\bar H$$.

Is this completely rigorous? I am not completely convinced that in the application of the correspondence theorem in the 2nd paragraph the image of $$H$$ under $$\pi$$ has to be $$\bar H$$. In the first paragraph the image of $$H$$ under the same map ostensibly has to be $$HN/N$$ for the theorem to make sense, but here it apparently is $$H/N$$. Is it because in the latter case $$H$$ contains $$N$$ and therefore $$HN=H$$? Assuming that's so, I am still not convinced that the image in the first paragraph has to be $$HN/N$$.

It is easy to check that $$\{hN: h\in H\}$$ equals to $$\{hN: h\in HN\}=HN/N$$. The reason why we don't write $$H/N$$ is because $$N$$ might not be a subgroup of $$H$$. But $$N\trianglelefteq HN$$ is true, so we can define the quotient group $$HN/N$$. And yes, you are right that if $$N\leq H$$ then $$HN=H$$, which is what you use in the second paragraph.