2
$\begingroup$

Suppose we have a simple finite group $G$ of odd order, and suppose $G$ has an automorphism $\alpha$ of order $2$ which centralizes precisely three elements of $G$. Let $H = \{ h \in G : h^{\alpha} = h^{-1} \}$ and put $T = C_G(\alpha)$. Then $T$ normalizes $H$.

Then we have $G = TH$.

I asked myself why this decomposition is valid. To show it I have to write every $g \in G$ as $g = th$ with $t \in T$ and $h \in H$. I see that for arbitrary $g \in G$, if we set $h := g^{-1}\alpha(g)$, then $\alpha(h) = \alpha(g)^{-1}g = h^{-1}$, so $h \in H$. But if we write $g = ghh^{-1}$, then $\alpha(gh) = \alpha(g)h^{-1} = g$, so this gives no such decomposition.

So why do we have $G = TH$?

Maybe this is of interest: If $H$ would be a subgroup, by results from this post it would be abelian.

By Feit-Thompson $G$ must be solvable, hence by simplicity $G' = 1$, so that $G$ must be abelian, and hence cyclic of prime power order. The only automorphism of order two would be inversion. So our group would not exist. But the above construction is part of a proof by contradiction (choosing a minimal counterexample), and in the reference it is explicitly mentioned by the authors that they do not want to use Feit-Thompson. So suppose that such a group exists, then why is $G = TH$ then?

$\endgroup$
  • $\begingroup$ Why the downvote? Is anything unclear or wrong with my question? $\endgroup$ – StefanH Jan 10 '16 at 23:07
  • 1
    $\begingroup$ By Feit-Thompson, a simple group of odd order is cyclic of prime order.... (In particular, the only automorphism of order $2$ is inversion, and it centralises only the identity, so your putative group does not exist.) $\endgroup$ – verret Jan 11 '16 at 1:22
  • $\begingroup$ @verret: Yes, thank you for pointing out! But if we assume that $G$ exists, why $G = TH$? I know this might be a little bit odd to ask, but it is part of a proof by contradiction I am trying to comprehend. Unfortunately the whole proof is quite long and uses other techniques, so posting the entire proof would be a little bit overkill. I added a paragraph explaining this. $\endgroup$ – StefanH Jan 11 '16 at 10:14
2
$\begingroup$

More generally, if $G$ is a finite group of odd order, $\alpha$ is an automorphism of $G$ of order $2$, $H = \{g \in G \mid \alpha(h)=h^{-1}\}$ and $T = C_G(\alpha)$, then $G=HT = TH$.

To see this let $K = \{ g^{-1}\alpha(g) \mid g \in G \}$. Then $K \le H$. Since $g^{-1}\alpha(g) = h^{-1}\alpha(h) \Leftrightarrow hg^{-1} \in T$, we have $|K| = |G|/|T|$.

So it is enough to show that distinct elements of $K$ lie in distinct left and right cosets of $T$, since this proves that $G=KT=TK$.

If not then (for right cosets - proof for left cosets is similar) $x=ty$ with $1 \ne t \in T$ and $x,y \in K$. Applying $\alpha$ gives $x^{-1}=ty^{-1}$, so $tyty^{-1} = 1$ and hence $t$ and $t^{-1}$ are conjugate in $G$, which is impossible when $|G|$ is odd.

(This argument applies to $x,y \in H$, so it proves that $|H| \le |G|/|T|$, so $K=H$.)

$\endgroup$
  • $\begingroup$ Thanks for your answer! But I guess in your 2nd paragraph you mean "$|K| = |G|/|T|$" instead of "$|K| = |G|/|H|$" as your bijection goes between $K$ and the cosets of $T$? Also in your 3rd paragraph I guess you mean "distinct left and right cosets of $T$" instead of "distinct left and right cosets of $H$", as this would imply $|K\cap T| = 1$ (or that $\bigcup_{k\in K} kT = G$), which I guess would imply $G = KT = TK$? $\endgroup$ – StefanH Jan 11 '16 at 13:40
  • 1
    $\begingroup$ Yes, I have corrected it! $\endgroup$ – Derek Holt Jan 11 '16 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.