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Let $N$ be a normal subgroup of $G$ and $\mathrm{Aut}(G)$ be the automorphism group of $G$.

Is there any example that $G/N$ is not isomorphic to $G/\gamma(N)$ for some $\gamma\in \mathrm{Aut}(G)$ ?

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    $\begingroup$ No. non όχι (I have to type a minimal number of characters!) $\endgroup$ – Bernard Feb 26 '15 at 21:41
  • $\begingroup$ @Bernard: Are you sure ? $\endgroup$ – mesel Feb 26 '15 at 21:43
  • $\begingroup$ Well, if $\varphi$ is the automorphism, $gN\mapsto \varphi(g)\varphi(N)$ is the isomorphism of $G/N$ onto $G/\varphi(N)$. $\endgroup$ – Bernard Feb 26 '15 at 21:49
  • $\begingroup$ @Bernard: The problem is that the map is not well defined. (take different repsesentative) $\endgroup$ – mesel Feb 26 '15 at 21:50
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    $\begingroup$ For me it is well defined: if $g^{-1}g'\in N$, $\varphi(g^{-1})\varphi(g')\in \varphi(N)$. Where is the problem? $\endgroup$ – Bernard Feb 26 '15 at 21:54
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$G / N$ is isomorphic to $G / \gamma(N)$ for any automorphism $\gamma$. The reason is that an automorphism $\gamma$ lifts to an isomorphism $\hat \gamma : G / N \rightarrow G / \gamma (N)$ where $\hat \gamma(x N) = \gamma (x) \gamma (N)$.

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