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

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 $$[a, G] = G$$.

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!

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 $$N_G(X_i)=N$$.

• Just a small nit: from the question it seems like $a\not\in G$, but is rather an automorphism of $G$. Of course your argument about normalizes applies to this case too. Commented Apr 22 at 20:00
• @SteveD Improved/fixed. Commented Apr 22 at 21:02
• Wow thank you so much for your detailed answer! This helped me a lot
– user1175180
Commented Apr 23 at 10:44
• Reading the proof again I have come up with one last unclearness. Why is $[X_i, N, \langle a \rangle] \leq [X_i, \langle a \rangle]$? Intuitively i would say it is becuase every $[x_i,n] \in [X_i,N]$ is an element of $X_i$ as well?
– user1175180
Commented Apr 23 at 16:53
• @Stippinator: $[x,y,z] = [[x,y],z]$. If $x\in X_i$ and $n\in N_G(X_i)$, then $[x,n] = x^{-1}(n^{-1}xn)\in X$, so $[X_i,N]\leq X_i$. Commented Apr 23 at 18:59