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  1. Let $H$ be a subgroup of $G$ such that $\varphi(H) \subseteq H$ for every automorphism. Show that $H \triangleleft G$.
  2. Let $Z(G)$ be the center of $G$. Show $\varphi(Z(G)) \subseteq Z(G)$ for all $\varphi \in \,Aut(G)$.

$\textbf{Show that $H \triangleleft G$.}$

I am having a hard time proving this part.

$\textbf{Show $\varphi(Z(G)) \subseteq Z(G)$ for all $\varphi \in \,Aut(G)$.}$

Well this is trivial from the previous proof.

Username Unknown
  • 4,633

3 Answers3

1

hints:

For (1): For all $\;g\in G\;$ , let $\;\phi_g\;$ be an inner automorphism, meaning:

$$\phi_gx:=x^g=g^{-1}xg$$

Apply now this to $\;H\;$ .

For (2) (which doesn't follow from (1), as noted in the comments): since for all $\;g\in G\;$ there exists $\;g'\in G\;$ s.t. $\;\phi(g')=g\;$ (why?) , we have that for all $\;z\in Z(G)\;$ :

$$g^{-1}\phi(z)g=\phi(g'^{-1})\phi(z)\phi(g')=\phi(g'^{-1}zg)=\ldots$$

Timbuc
  • 34,191
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1) Since $\phi(H)\subset H$ $\forall\ \phi \in Aut(G)$, hence in particular this will hold for the $inner\ automorphisms$ of $G$, i.e. $\forall\ g\in G$, $T_g(x)=gxg^{-1}$. Thus $\forall\ g\in G$ $T_g(H)=gHg^{-1}\subset H$ $\Rightarrow$ $H\unlhd G$.

2) Hint :- Let $x\in G$, $y\in G$ and $\phi$ be any automorphism of $G$. Show that $\phi(x)\phi(y)=\phi(y)\phi(x)$. Thus proving that $\phi(x)$ commutes with every element of $G$ and hence $\phi(Z(G))\subset Z(G)$.

Note :- 2) does $not$ follow from 1) as you have said.

wanderer
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Let us fix an element $g\in G$

Let $\phi:G\rightarrow G $ be defined by $\phi (h)=ghg^{-1}$Clearly an automorphism from $G$ to $G$.Then $\phi(H)\subseteq H$ .Thus $ H$ is normal in $G$

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