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A group is called a $C_\pi$-group if there exists a Hall $\pi$-subgroup and any two Hall $\pi$-subgroups are conjugate.

Let $G$ be a group such that $N$ is a normal $C_\pi$-subgroup of $G$ and suppose that $G/N$ is a $C_\pi$-group. Then $G$ is a $C_\pi$-group.

I have already shown the existence of a Hall $\pi$-subgroup of $G$ and I only need to prove the conjugacy part.

Suppose that $H$ and $K$ are Hall $\pi$-subgroups of $G$. Then $HN/N$ and $KN/N$ are Hall $\pi$-subgroup of $G/N$. Then there exists $x\in G$ such that $HN = xKx^{-1}N$

Also $H\cap N$ and $xKx^{-1}\cap N$ are Hall $\pi$-subgroup of $N$ and since N satisfies $C_\pi$, there exists $y\in N$ such that $H \cap N = yxKx^{-1}y^{-1} \cap N$

Let $L=HN$. Then $N_L(H\cap N)$ contains both $H$ and $yxKx^{-1}y^{-1}$

Consider the factor group $N_L(H \cap N)/H \cap N$. Since $N_N(H \cap N) \unlhd N_L(H \cap N)$, we have that $N_N(H \cap N)/H \cap N \unlhd N_L(H \cap N)/ H \cap N$.

Also $[N : H\cap N]= [N : N_N(H \cap N)] [N_N(H \cap N) : H\cap N]$ and since $[N : H\cap N]$ is a $\pi'$-number, we have $[N_N(H \cap N) : H\cap N]$ is a $\pi'$-number.

I need to show that $[N_L(H \cap N) : N_N(H\cap N)]$ is a $\pi$-number so I can use the Schur-Zassenhaus theorem, but I'm unable to do this so far

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  • $\begingroup$ Here is hint. Prove that $L = N_L(H \cap N)N$. The proof is essentially the Frattini Argument. Then $N_L(H \cap N)/N_N(H \cap N) \cong L/N$, which is a $\pi$-group. $\endgroup$ – Derek Holt May 17 '16 at 19:55
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I know that $H\leq N_L(H\cap N) \leq L$. Hence $L = HN \leq N_L(H\cap N)N$ and thus implying $L = N_L(H\cap N)N$

Now $|HN/N|=|L/N|$. Also $|L/N| = \dfrac{|N_L(H\cap N)|}{|N_L(H\cap N) \cap N|}$

Clearly, $N_L(H\cap N) \cap N \leq N_N(H \cap N) \leq N_L(H\cap N) $

$[L:N] = [N_L(H\cap N) :N_N(H\cap N)][N_N(H\cap N) : N_L(H\cap N) \cap N]$

Since $[N_L(H\cap N) :N_N(H\cap N)]$ divides a $\pi$-number, it must be a $\pi$-number itself.

Does this prove what you were saying?

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  • $\begingroup$ Yes that looks OK. $\endgroup$ – Derek Holt May 18 '16 at 8:06

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