$G$ is a finite group, $\alpha$ is an automorphism of $G$ and $I=\{g\in G\mid\alpha(g)=g^{-1}\}$. If $|I|=\frac34|G|$, show that $G$ has an abelian subgroup of index 2.
I don't know where to start.
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spiritfire
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1You could start by trying to adapt the proof of the related question and see how far you get. You get $|T| \ge |G|/2$ so either $G$ is abelian, and you are OK, or $T = C_G(s)$ has order $|G|/2$ for some $s \in S$. Then note that $\alpha$ fixes the subgroup $C_G(s)$. – Derek Holt Jul 25 '16 at 15:17
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@Derek Holt: If everything in the proof of the related question goes through, then I can follow your sketch and prove that T is abelian, but I don't know why $|T_2|\lt \frac {|G|}{4}$ in the proof of $|T|\ge \frac {|G|}{2}$. – spiritfire Jul 26 '16 at 05:38
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You have $|T_1|=|T_2|=|G|/4$, giving $|T| \ge |G|/2$. – Derek Holt Jul 26 '16 at 07:46
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@Derek Holt: Why$ |T_1|=|T_2|$? – spiritfire Jul 26 '16 at 09:05
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Because they are both equal to $|G|-|S|$ (i.e. $|G|-|I|$ in your notation). – Derek Holt Jul 26 '16 at 11:34
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Why $|T_2|$=|G|-|S|? I can't find an bijection between $T_1$ and $T_2$, nor find $|T_2|$ more directly. – spiritfire Jul 26 '16 at 12:56
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And in your first comment, that $\alpha$ fixes the subgroup $C_G(s)$ doesn't help my proof that T is abelian. – spiritfire Jul 26 '16 at 12:59
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There is a very obvious bijection between $T_2$ and $G \setminus S = T_1$. The fact that $\alpha$ fixes $T = C_G(s)$ and inverts each of its elements implies that it is abelian. – Derek Holt Jul 26 '16 at 13:30
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Let us continue this discussion in chat. – spiritfire Jul 26 '16 at 13:39
1 Answers
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Hint: if $a,b\in I$ and $ab\in I$ then we have $ab=ba$.
1.We can claim that $I\cap hI\subset C(h)$ for any $h\in I$.
If $x\in I\cap hI$ then $x=hg$ with $h,g\in I$. $\alpha(gh)=(gh)^{-1}$ since $x\in I$; on the other hand, $\alpha(gh)=\alpha(g)\alpha(h)=g^{-1}h^{-1}$. Hence $(gh)^{-1}=g^{-1}h^{-1}=(hg)^{-1}$, this is, $gh=hg$. Thus $g\in C(h)$ and $x=hg\in C(h)$ since $C(h)$ is a subgroup.
2.If $|I|=\frac{3}{4}|G|$ then $G$ can not be abelian.(if $G$ is abelian then $I$ is a subgroup of $G$ which contradicts $|I|=\frac{3}{4}|G|$)
Hence there exists $h\in I-C(G)$ and $|C(h)|<|G|$.
$|I\cap hI|=|I|+|hI|-|I\cup hI|\ge \frac{1}{2}|G|$ and $|C(h)|\ge \frac{1}{2}|G|$.
Hence $|C(h)|=\frac{1}{2}|G|$ and $C(h)=I\cap hI$.
3.The only thing left to show is that $C(h)$ is abelian.
For any $k=hg\in C(h)$ then we have $g=h^{-1}k\in C(h)$. For $k_1=hg_1,k_2=hg_2$, we have $g_1g_2\in C(h)\subset I$. Then $(g_1g_2)^{-1}=\alpha(g_1g_2)=\alpha(g_1)\alpha(g_2)=g_1^{-1}g_2^{-1}$ and $g_1$ commutes $g_2$. Thus $k_1k_2=hg_1hg_2=h(hg_1)g_2=h^2(g_2g_1)=h(g_2h)g_1=k_2k_1$.
Hence $C(h)$ is an abelian subgroup of index $2$.
wzstrong
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