# Question on the second homomorphism theorem.

I'm not much familiar with this topic, but nontheless I have a problem understanding the context. Following this link Second Isomorphism Theorem, they state that $$HN/N$$ is a group under left coset multiplication. They then define the map $$\phi(h)=hN$$ (a map from $$H$$ to $$HN/N$$) where $$N$$ is a normal subgroup of a larger group $$G$$ and $$H$$ is a subgroup of $$G$$. Furthermore they state $$\phi$$ is clearly a homomorphism. How is that? How do I understand the algebraic operation between these elements?

For instance, we would have to have $$\phi(h\cdot h^{-1})=\phi(e)=N=\phi(h)\cdot \phi(h)^{-1}=hN\cdot (hN)^{-1}$$. So as far as I undertand, the elements of $$HN/N$$ are the cosets of $$HN$$, right? How do you take the inverse of a coset?

• Yes, I see that, but why is $NN=N$? What is $NN$ even supposed to mean? Jun 1 at 11:55
• Sorry, I just deleted my comment to write an answer :-) $NN=\{nm\mid n, m\in N\}$ and by taking $n=1$ we see that $N\leq NN$, while clearly $NN\leq N$ as $N$ is closed under multiplication. Jun 1 at 11:57
• Isnt' $\varphi$ the restriction to H of the projection on the quotient? Jun 1 at 11:58
• @Diger $NN=\{nm | n\in N, m\in N\}$ Jun 1 at 12:00
• Is it just a definition, or does it follow from a specific operation? Jun 1 at 12:00

I think the thing you are struggling with is the definition of the group operation on $$HN/N$$, or more generally on $$G/N$$. Here, multiplication is defined as: $$gNhN=ghN$$ which makes sense, as by normality of $$N$$ we have $$gNhN=g(hNh^{-1})hN=ghNN=ghN$$. Therefore, inverses are as follows: $$(gN)^{-1}=g^{-1}N$$

As the map you are considering is defined by $$\phi(h)=hN$$, we clearly have: $$\phi(g)\phi(h)=gNhN=ghN=\phi(gh)$$ as required for homomorphisms.

• So, just for completeness: I could think of $NN^{-1}=\{nm^{-1}|n,m\in N\}=\{nm|n,m\in N\}=N$, since $N$ is a group $m^{-1}\in N$ and when $m$ runs over all elements of $N$, so does $m^{-1}$. Jun 1 at 12:15
• Yeah - $NN=N=N^{-1}N=NN^{-1}$ etc. Jun 1 at 12:23

The "elements" in the quotient group are those cosets of $$N$$, and $$N$$ works as the unit element. Suppose you take some element $$hN$$, the inverse is also some coset, say $$bN$$, which satisfies

$$hNbN=N$$

Since $$N$$ is normal, namely $$Nb=bN$$, hence we have

$$hNbN=hbNN=hbN=N$$

This implies

$$hb\in N\Longrightarrow b\in h^{-1}N\Longrightarrow b=h^{-1}n_1,~~n_1\in N$$

so we get the inverse

$$(hN)^{-1}=bN=h^{-1}n_1N=h^{-1}N$$

Consider a group $$G$$, $$H$$ subgroup of $$G$$ and $$N$$ a normal subgroup of $$G$$. The projection $$p:G\to G/N$$ defined letting $$p(g)=gN$$ for all $$g\in G$$ is well defined and it's a surjective group homomorphism. It's easy to prove that $$p(H)=HN/N$$, hence $$\phi=(p_{|H})^{|HN/N}$$ is a group homomorphism.