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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?

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  • $\begingroup$ Yes, I see that, but why is $NN=N$? What is $NN$ even supposed to mean? $\endgroup$
    – Diger
    Commented Jun 1, 2023 at 11:55
  • $\begingroup$ 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. $\endgroup$
    – user1729
    Commented Jun 1, 2023 at 11:57
  • $\begingroup$ Isnt' $\varphi$ the restriction to H of the projection on the quotient? $\endgroup$ Commented Jun 1, 2023 at 11:58
  • $\begingroup$ @Diger $NN=\{nm | n\in N, m\in N\}$ $\endgroup$ Commented Jun 1, 2023 at 12:00
  • $\begingroup$ Is it just a definition, or does it follow from a specific operation? $\endgroup$
    – Diger
    Commented Jun 1, 2023 at 12:00

3 Answers 3

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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.

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  • $\begingroup$ 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 $m^{-1}$. $\endgroup$
    – Diger
    Commented Jun 1, 2023 at 12:15
  • $\begingroup$ Yeah - $NN=N=N^{-1}N=NN^{-1}$ etc. $\endgroup$
    – user1729
    Commented Jun 1, 2023 at 12:23
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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

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

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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.

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