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Please critique my proof:

Let $f: G \to H$ be a group homomorphism and let $U \leqslant G.$ Show that $f^{-1}(f(U)) = U \cdot \ker(f).$

$(i) \hspace{1mm} f^{-1}(f(U)) \subseteq U \cdot \ker(f):$ Let $a \in f^{-1}(f(U)).$ Then $f(a) \in f(U) \Rightarrow$ There exists $u \in U$ such that $f(u) = f(a).$ Let $v = u^{-1} a$, then $a = u v$. This shows $v \in \ker(f)$ since $f(v) = f(u^{-1} a) = f(u^{-1}) f(a) = f(u)^{-1} f(a) = f(u)^{-1} f(u) = 1_H.$ Thus $a = uv \in U \cdot \ker(f).$

$(ii) \hspace{1mm} f^{-1}(f(U)) \supseteq U \cdot \ker(f):$ Now assume that $a \in U \cdot \ker(f).$ Then $a = uu'$ for $u \in U$ and $u' \in \ker(f).$ Thus $f(uu') = f(u)1_H = f(u) \in f(U)$ and so $uu' = a \in f^{-1}(f(U)).$ Combining $(i)$ and $(ii)$ completes the proof.

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    $\begingroup$ In (i) why is $a\in U$? $\endgroup$
    – scineram
    Mar 16, 2015 at 5:26
  • $\begingroup$ I think I made a mistake, now I am not sure how to continue. $\endgroup$
    – St Vincent
    Mar 16, 2015 at 6:59

1 Answer 1

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There exists $u \in U$ such that $f(u) = f(a).$ Let $v = u^{-1} a$, then $a = u v$. $f(v) = f(u^{-1} a) = f(u^{-1}) f(a) = f(u)^{-1} f(a) = 1_H$. So $v \in \ker(f)$.

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