If $X$ is a subnormal in $G$ with $[a, X] = 1 = [a, C_G(X)]$, then $[a, G] = G$

Question

Let $G$ be a $\pi$-group and $a$ $\pi'$-element that acts on $G$. If $X$ is a subnormal subgroup of $G$ with $[a, X] = 1 = [a, C_G(X)]$, then $[a, G] = G$.

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Proof
Let $X = X_0 \unlhd X_1 \unlhd \ldots \unlhd X_n = G$ be a subnormal series. Choose $i$ maximally such that $[a,X_i]=1$ and set $N=N_G(X_i)$. Then $a$ also acts on $N$.

Assume $i \neq n$. Since $X_{i+1} \leq N$, it follows that $[a,N] \neq 1$. We have $$[X_i, N, \langle a \rangle] \leq [X_i, \langle a \rangle] = 1 \text{ and }[\langle a \rangle, X_i, N] = 1.$$ By the Three Subgroups Lemma, it is also $[N, \langle a\rangle, X_i]=1$. Thus, $[N,a]\leq C_G(X_i) \leq C_G(X)$. Since $[C_G(X), \langle a \rangle] = 1$, it follows $[N, \langle a \rangle, \langle a \rangle] = 1$ and by Lemma 2.2.10([G,A,A] = [G,A] for coprime actions), it then follows $[N, \langle a \rangle]=1$, a contradiction. Therefore, $i=n$ and $[G,a]=1$.
q.e.d.

Hello everyone!

I'm having trouble with a couple of concepts from a group theory proof we discussed in class and hoped somone could walk me through there. Specifically, I'm not clear on why $a$ acts on the defined subgroup $N$ and why $X_{i+1} \leq N$. This makes it also hard for me to understand the remaining equations. Could anyone help clarify these points for me?

Thanks a lot for your help!

Answer 1

For any subgroup $H$ of a group $G$ and any $\rho$ that acts on $G$ (I will have them act on the left, but you can have them act on either side), we have that $ {}^{\rho}N_G(H) = N_G({}^{\rho}H)$. To see this, note that if $x\in N_G(H)$, and $h\in H$, then $${}^{\rho}x({}^{\rho}h){}^{\rho}x^{-1} = {}^{\rho}(xhx^{-1})\in {}^{\rho}H,$$ since $xhx^{-1}\in H$. Thus, ${}^{\rho}N_G(H)\subseteq N_G({}^{\rho}H)$. Applying this argument to ${}^{\rho}H$ and $\rho ^{-1}$, we get $$ {}^{\rho^{-1}}N_G({}^{\rho}H)\subseteq N_G({}^{\rho^{-1}}{}^{\rho}H) = N_G(H),$$ and applying $\rho$ to both sides we get $N_G({}^{\rho}H) \subseteq {}^{\rho}N_G(H)$, giving equality.

Now, you have $[a,X_i]=1$, so $a$ fixes every element of $X_i$. By the above, we have $${}^aN = {}^aN_G(X_i) = N_G({}^aX_i) = N_G(X_i)=N,$$ so $a$ acts on $N$.

Since $X_i$ is normal in $X_{i+1}$, every element of $X_{i+1}$ normalizes $X_i$, and therefore $X_{i+1}$ is contained in $N_G(X_i)=N$.