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  1. #1
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    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 \pi: G \rightarrow G/N is the natural projection homomorphism of G on G/N and \Phi: G \rightarrow H is a homomorphism, then let's prove that \phi, given by gN \mapsto \Phi(g) is a function and homomorphism if and only if  N \subseteq ker \Phi.

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

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

    Thanks in advance....
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  2. #2
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    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").
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  3. #3
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    Re: Question about induced homomorphism

    Thanks, Deveno!
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