Well-defined: We need to show that if $x_1 \in x$ and $x_2 \in x$, then $f(x_1) = f(x_2)$. Since $x_1, x_2 \in x$, they are in the same coset of $\ker f$, which means $x_1^{-1}x_2 \in \ker f$. Thus, $f(x_1^{-1}x_2) = e_{G'}$. Now, we have $f(x_1)^{-1}f(x_2) =
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