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I am trying to understand this theorem from Munkres:

If $G$ is the free product $G_1*G_2$, and $N_1$,$N_2$ are some normal subgroups of $G_1$ and $G_2$, respectively, and $N$ is the least normal subgroup of $G$ containing $N_1$ and $N_2$, then

$G/N \cong (G_1/N_1)*(G_2/N_2)$

I cannot convince myself that $G-> (G_1*G2)/N$ will map $N_1$ to identity. Can anyone see why that holds ?

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Because $N_1$ is contained in $N$ by definition. –  anon Mar 30 '13 at 21:08
    
Oh wow that is embarrassing. For some reason I kept trying to prove that it is mapping G to identity. Nevermind! Turns out just typing out stupid questions helps me think better. –  user66306 Mar 30 '13 at 21:12

1 Answer 1

up vote 1 down vote accepted

$N_1$ is contained in $N$, so since $G \to G/N$ maps $N$ to identity, it also sends $N_1$ to identity.

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