1. ## Question about induced homomorphism

Greetings,

I have problem understanding something I read of Dummit & Foote. I'm not sure I got it correctly, so I need to be checked:

If $\displaystyle \pi: G \rightarrow G/N$ is the natural projection homomorphism of $\displaystyle G$ on $\displaystyle G/N$ and $\displaystyle \Phi: G \rightarrow H$ is a homomorphism, then let's prove that $\displaystyle \phi$, given by $\displaystyle gN \mapsto \Phi(g)$ is a function and homomorphism if and only if $\displaystyle N \subseteq ker \Phi$.

If I understand correctly, the issue here is proving that $\displaystyle \phi$ is a function. If we know that $\displaystyle \phi$ is a function, then proving that
$\displaystyle \phi(g_1N)\phi(g_2N)=\phi(g_1g_2N)$ is easy: $\displaystyle \phi(g_1N)=\Phi(g_1)$,$\displaystyle \phi(g_2N)=\Phi(g_2)$,$\displaystyle \phi(g_1g_2N)=\Phi(g_1g_2)$ and $\displaystyle \Phi(g_1)\Phi(g_2)=\Phi(g_1g_2)$ because $\displaystyle \Phi$ is homomorphism.

So, assume $\displaystyle N \subseteq ker \Phi$ and let $\displaystyle g_1N=g_2N$ (we want to prove that this implies $\displaystyle \phi(g_1N)=\phi(g_2N)$). Then $\displaystyle g_1$ and $\displaystyle g_2$ are in the same coset of N and this coset is contained in one coset of $\displaystyle ker\Phi$, because $\displaystyle gN \subseteq g ker\Phi$ for any $\displaystyle g$. So $\displaystyle g_1$ and $\displaystyle g_2$ must be in the same coset of $\displaystyle ker\Phi$, so $\displaystyle \Phi(g_1)=\Phi(g_2)$ which means that $\displaystyle \phi(g_1N)=\phi(g_2N)$.

And I'm so far. I would like to know if the above checks up ok and how to prove the reverse statement.

2. ## Re: Question about induced homomorphism

Yes the issue is whether or not $\phi$ is well-defined.

If we assume $\phi$ IS well-defined, this means that $\phi$ is constant on any coset of $N$ (it has to be, or else $\Phi$ is not a function).

In particular, $\phi$ must be constant on the coset of the identity, $N$. That is, for any $n \in N, \phi(N) = \phi(nN) = \Phi(n)$. Since $e \in N$, we must have $\phi(N) = \phi(eN) = \Phi(e)$, which shows that $\Phi(n) = \Phi(e) = e_H$

that is: $N \subseteq \text{ker }\Phi$.

One way to do this without "element-chasing" (using the $g_1,g_2$) is to say $\Phi$ factors through $\pi$, that is, assert there is a homomorphism $\phi$ such that: $\Phi = \phi\pi$. What we have just shown is that this happens iff $\Phi$ annihilates $N$.

The "canonical nature" of $\pi$ is then expressed as: $\pi$ is UNIVERSAL among homomorphisms that annihilate $N$.

(The term "annihilate" is used because the trivial group is a "zero object" for the class of all groups, in the sense that there is always UNIQUE homomorphisms:

$\iota: \{e\} \to G$ <--this says the trivial group is an "initial group"
$\tau: G \to \{e\}$ <--this says the trivial group is a "terminal group"

for any group $G$.

In other words, a trivial group acts much the same as a trivial vector space (consisting just of a 0-vector), which has similar properties with "linear mapping" substituted for "homomorphism").

3. ## Re: Question about induced homomorphism

Thanks, Deveno!