Let $\phi:G\to G'$ be a group homomorphism, and let $N'$ be a normal subgroup of $G'$. Show that $\phi^{-1}[N']$ is a normal subset of $G'$.

My attempt: $\phi^{-1}[N']=\{g\in G:\phi(n)\in N'\}$ $$g' n' (g')^{-1}\in N'$$ $$\Rightarrow\phi^{-1}(g' n' (g')^{-1})=\phi^{-1}(g')\phi^{-1}(n')\phi^{-1}(g'^{-1})$$ Thus $\phi^{-1}[N']$ is normal. Is this thinking correct? I feel like I am missing something.

  • $\begingroup$ I don't know if this is just happening on my phone, but the \prime command looks terrible. Normally it is sufficient to use an apostrophe. $\endgroup$ – Matt Samuel Nov 3 '15 at 4:41
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    $\begingroup$ @MattSamuel One can use \prime but it needs to be a superscript, like $g^\prime$. Of course $g'$ is simpler to type, and is the shorthand for the former. $\endgroup$ – user147263 Nov 3 '15 at 4:42

Step $1$: Show that $\phi^{-1}[N]\neq \emptyset $

Step $2$: Take $g_1,g_2\in \phi^{-1}[N]$ show that $g_1g_2\in \phi^{-1}[N]$

Step $3$:Take $g\in G,x\in \phi^{-1}[N]$ show that $gxg^{-1} \in \phi^{-1}[N]$


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