On group theory 11

[Papapetros Vaggelis]

Let \(\displaystyle{\left(G,\cdot\right)}\) be a finite group with \(\displaystyle{|G|=2\,n+1\,,n\in\mathbb{N}}\)

and \(\displaystyle{f\in\rm{Aut}(G)}\) such that \(\displaystyle{f^2=Id_{G}}\) .

Prove that for every prime number \(\displaystyle{p\mid |G|}\), there exists a \(\displaystyle{P}\)

\(\displaystyle{\rm{Sylow}-p}\) subgroup of \(\displaystyle{\left(G,\cdot\right)}\) such that \(\displaystyle{f(P)=P}\) .

[dmsx]

By the Sylow theorems, Sylow $p$-subgroups exist, since $p$ divides $|G|$. We observe that for every $P \in Syl_p(G)$, we have $f(P) \in Syl_p(G)$, since automorphisms preserve orders.

*If there is only one Sylow $p$-subgroup, call it $P$, we have $f(P)=P$.

*In the general case, let $P_1,...,P_m$ be the (distinct) Sylow $p$-subgroups of $G$. Pick a Sylow $p$-subgroup $P_i$. As we noticed earlier, $f(P_i)$ is also a Sylow $p$-subgroup of $G$, so it is contained in the conjugate of another Sylow $p$-subgroup, which is again a Sylow $p$-subgroup by the Sylow theorems. Call that one $P_j$, where $j \in \left \{ 1,...,m \right \}$. So $f(P_i)\subseteq {P_j}$ and because $f \in Aut(G)$, we have $f(P_i)={P_j}$. If $i=j$, we are done. If not, by applying $f$ once more, we get $f^2(P_i)= f({P_j})\Rightarrow P_i=f({P_j})$, since $f^2 \equiv id_G$.

This means that every two Sylow $p$-subgroups $P_i, P_j$, for $i\neq j$, are matched in this way. Now, since $n_p=m$ and $n_p$ divides $|G|$ which is odd, we deduce that $m$ is also odd. That means there is an odd number of Sylow $p$-subgroups. Considering the pairing we described, we can find an index $k$, so that the Sylow $p$-subgroup $P_k$ is equal to its image under $f$, so we have $f(P_k)=P_k$.

[Papapetros Vaggelis]

Hi dmsx. Thank you for your solution.

Here is another one.

Solution

Let \(\displaystyle{p}\) be a prime number such that \(\displaystyle{p\mid |G|}\) . According to

\(\displaystyle{\rm{Sylow}}\) theorems, we have that \(\displaystyle{\rm{Sylow}-p}\) subgroups exist.

We consider the set \(\displaystyle{X}\) of all the \(\displaystyle{\rm{Sylow}-p}\) subgroups and then

\(\displaystyle{X\neq \varnothing}\). We define a left action

\(\displaystyle{\star:\langle{f\rangle}\times X\longrightarrow X\,\,,Id_{X}\star P=P\,\,,f\star P=f(P)\,\,,\forall\,P\in X}\).

\(\displaystyle{\forall\,P\in X: [P]_{\langle{f\rangle}}=\left\{P,f(P)\right\}}\). We have that

\(\displaystyle{X=\bigcup_{P\in S}[P]_{\langle{f\rangle}}}\), where \(\displaystyle{S}\) is a set

of representatives of orbits of the above action. So,

\(\displaystyle{n_{p}=|X|=\sum_{P\in S}|[P]_{\langle{f\rangle}}|}\). Since \(\displaystyle{n_{p}\mid |G|}\)

and \(\displaystyle{|G|}\) is an odd number, we get that \(\displaystyle{n_{p}}\) is an odd number.

If \(\displaystyle{|[P]_{\langle{f\rangle}}|=2\,,\forall\, P\in S}\), then \(\displaystyle{n_{p}}\) is

an even number, a contradiction, so there exists \(\displaystyle{P\in S\subseteq X}\) such that

\(\displaystyle{|[P]_{\langle{f\rangle}}|=1}\), that is \(\displaystyle{f(P)=P}\) .