1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Nov 3, 2015 at 4:41
  • 1
    $\begingroup$ @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, 2015 at 4:42

2 Answers 2

2
$\begingroup$

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 $gxg^{-1} \in \phi^{-1}[N]$

$\endgroup$
1
$\begingroup$

Fleshing out @Learnmore's answer:

Step 1:

Note that $N'\unlhd G'$ implies $\phi(e_G)=e_{G'}=e_{N'}$. Thus $e_G\in\phi^{-1}(N')$, so $\phi^{-1}(N')\neq \varnothing$.

Step 2:

Let $g,h\in \phi^{-1}(N')$. Then there exist $m,n\in N'$ with $\phi(g)=m,\phi(h)=n$. Now

$$\begin{align} \phi(gh)&=\phi(h)\phi(h)\\ &=mn\\ &\in N' \end{align}$$

since $N'$ is a subgroup of $G'$. Therefore, $gh\in\phi^{-1}(N')$.

Step 3:

Let $g\in G, x\in\phi^{-1}(N').$ Then there is some $r\in N'$ with $\phi(x)=r$. We have

$$\begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)\phi(x)\phi(g)^{-1}\\ &=\phi(g)r\phi(g)^{-1}\\ &=g'rg'^{-1}, \end{align}$$

where $g'=\phi(g)\in G'$, which implies $g'rg'^{-1}\in N'$ as $N'\unlhd G'$. Hence $gxg^{-1}\in\phi^{-1}(N')$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .