Let $G$ be a finite group and $U \le G$ be a finite group of odd order. Suppose that $N_G(U) = TU$ where $T = \langle t \rangle$ for some involution $t \notin U$. Also suppose $U^g \ne U$ implies $U^g \cap U = 1$.

Assume $M \unlhd G$ such that $M \cap U = U'$ and $G' \le M$. Then $M/G'$ is a Hall-subgroup of $G/G'$.

How to show that? We have that $M/G'$ is a Hall-subgroup iff $\operatorname{gcd}(|M : G'|, |G : M|) = 1$. I guess the best might be to show that every Sylow subgroup of $M/G'$ is also a Sylow subgroup of $G / G'$, but here I am stuck and do not know how to proceed. So any help would be appreciated!?

  • $\begingroup$ What are the conditions on $G'$? Is it just a normal subgroup of $G$ that is contained in $M$? If so, then it is enough to show for $G'=\{e\}$, but then you have no condition on $M$ other than it is normal in $G$. In any way, it seems that the group $G'$ is irrelevant for the proof. $\endgroup$ – Ofir Jan 26 '16 at 13:22
  • $\begingroup$ @Prometheus: $G'$ (and similar $U'$) is the commutator subgroups, $G' = [G, G] = \langle [x,y] : x,y \in G \rangle$, and similar $U'$ is the group generated by all commutators made by elements from $U$. $\endgroup$ – StefanH Jan 26 '16 at 15:25
  • $\begingroup$ ok, that makes more sense $\endgroup$ – Ofir Jan 26 '16 at 16:53
  • 2
    $\begingroup$ If $P$ is a nontrivial Sylow $p$-subgroup of $U$ then $P \in {\rm Syl}_p(G)$ and by the Frattini Argument $G=N_G(P)UG'$. But $|N_G(P):N_U(P)| = 2$, so $|M/G'| \le 2$. So $M/G'$ has order $1$ or $2$, and it is a Hall subgroup of $G/G'$ but in a rather trivial fashion. $\endgroup$ – Derek Holt Jan 27 '16 at 9:15
  • $\begingroup$ @DerekHolt Thanks! Now I solved everything. But the computations to get to $|M/G'| \le 2$ were quite messy (see my answer at iii)), if you have any easier argument than mine I would be glad if you share it! $\endgroup$ – StefanH Jan 27 '16 at 13:17

With Derek's comment I was able to solve it:

i) If $P \in \mbox{Syl}_p(U)$, then $P \in \mbox{Syl}_p(G)$.

Assume $P < Q$ for some Sylow $p$-subgroup of $G$, then as $P < N_Q(P) \le Q$ we can choose $g \in N_Q(P) \setminus P$. Then $P = P \cap P^g \le U \cap U^g$, hence as $P \ne 1$ we have $U = U^g$ and therefore $g \in TU$. As $g \notin U$ we can assume $g = tu$, but $tu$ has even order as $(tu)^2 \in U$.

A different way of seeing this: Consider the transitive action of $G$ on the right cosets of $U$ in $G$. As $Uxg = Ux \Leftrightarrow g \in U^x$ every $g \in G$ fixes at most two cosets. For assume $g \in U^x \cap U^y \cap U^z$ where $|\{U^x, U^y, U^z\}| = 3$, then $g^{x^{-1}} \in U \cap U^{yx^{-1}}$, hence $U = U^{yx^{-1}}$ and similar $U = U^{zx^{-1}}$. But $|\{ U, Uyx^{-1}, Uzx^{-1} \}| = 3$, and hence we would have three different cosets in $N_G(U)$, but this is not possible.

To see that the cosets are different, note that $Ux = Uy \Leftrightarrow yx^{-1} \in U$ implies $U^x = (U^{yx^{-1}})^x = U^y$ (or use $Ux = Uy \Leftrightarrow x^{-1}U = y^{-1}U$). So for example $U = Uyx^{-1}$ would give $U^x = U^y$ and similar for the others.

Now look at the orbits of $P$ on the cosets of $U$, as $P$ could fix at most two cosets, we have with $|P| = p^k$ $$ |G : U| \equiv 1 \pmod{p} \quad\mbox{ or }\quad |G : U| \equiv 2 \pmod{p} $$ and in both cases $|G : U|$ is not divisible by $p > 2$.

Note that by the equivalence of group actions and actions on cosets of some subgroup, in above terms we have shown that for three different points (the cosets), the point stabilizers corresponding to these points intersect trivially (the point stabilizers are the conjugates of $U$, as $G_{Ux} = (G_U)^x = U^x$ with common notation for stabilizers). Also note that by the condition that an element has just one fixed point is also excluded, as if $g \in U^x$, then $g^{x^{-1}} \in U = U^t$, hence $g^{x^{-1}}$ fixes also $Ut$.

ii) $G = N_G(P)UG'$ and $|N_G(P) : N_U(P)| = 2$.

As $G' \le UG'$ the subgroup $UG'$ is normal, and by i) in particular $P$ is also a Sylow subgroup of $UG'$, hence Frattini gives the first claim. As $1 \ne P\cap P^g$ implies $g \in N_G(U)$ we have $N_G(P) \le TU$. Now if $N_U(P) = U$, then $P$, as a characteristic subgroup in $U$, is normal in $TU$, hence $TU \le N_G(P)$. Otherwise we have some $u \in U$ not normalising $P$, which gives $UN_G(P) = TU$. Further $2 = |TU : U| \ge |TU \cap N_G(P) : U \cap N_G(P)| = |N_G(P) : N_U(P)|$ and by the remarks before we have $N_U(P) \ne N_G(P)$.

iii) $|M / G'| \le 2$ and $M/G'$ is a Hall subgroup in $G/G'$.

By similar arguments as in i) we have $G = UM$, hence $$ G / M \cong UM / M \cong U / (U \cap M) \cong U / U' $$ and so $G/M$ is abelian, which gives $G' \le M$ (so this assumption could be dropped in the question) and $|G : M|$ is odd as $|U|$ has odd order.

Now $$ |M : U'| = |M : G'|\cdot |G' : U'| $$ and \begin{align*} |M : U'| & = |M : M\cap U| \\ & = |MU : U| \\ & \le |G : U| \\ & = |N_G(P)UG' : U| \\ & = |N_G(P)UG' : N_G(P)U||N_G(P)U : U| \\ & = |G' : N_G(P)U \cap G'||N_G(P)U : U| \\ & = |G' : N_G(P)U \cap G'||N_G(P) : N_G(P) \cap U| \\ & = 2 \cdot |G' : N_G(P)U \cap G'| \\ \end{align*} and we have $U' \le N_G(P)U \cap G'$ so that $|G' : N_G(P)U \cap G'| \le |G : U'|$. Taken together $$ |M : G'||G' : U'| \le 2|G' : U'| $$ hence $|M : G'| \le 2$ and $|G/G' : M/G'| = |G:M|$ has odd order.


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