# $G = \langle C_G(a) \mid a \in Q \setminus \{1\} \rangle.$ Proof by induction on $|G|$

I have a question on a proof found in our lecture book on group theory from Gernot Stroth. I do not understand what here is meant by "Induction on $$|G|$$", which makes it very hard to grasp the last two equations. It would be very helpful if someone could give me a hint.

Lemma 2.2.3 Let $$p, q$$ be distinct prime numbers, $$G$$ an abelian $$p$$-group, and $$Q$$ a non-cyclic abelian $$q$$-group of automorphisms of $$G$$. Then $$G = \langle C_G(a) \mid a \in Q \setminus \{1\} \rangle.$$

Proof: We prove the claim by induction on $$|G|$$.

First, suppose $$C_G(x) = 1$$ for all $$x \in Q \setminus \{1\}$$. We have a subgroup $$Q_1$$ of $$Q$$ with $$|Q_1| = q$$. Since $$Q$$ is abelian, there must be another subgroup $$Q_2$$ of $$G$$ such that $$Q_1 \neq Q_2$$ and $$|Q_2| = q$$. However, $$Q_1 Q_2 = Q_1 \times Q_2$$ is not cyclic, which contradicts Lemma 2.2.2.

So, there exists $$x \in Q \setminus \{1\}$$ with $$C_G(x) \neq 1$$. By Lemma 2.2.1(b), we have $$G = [G, x] \times C_G(x)$$. Therefore, $$|[x, G]| < |G|$$. For all $$y \in Q$$, we have $$[x, G]^y = [x^y, G^y] = [x, G]$$. Thus, $$Q$$ acts on $$[x, G]$$. By induction on $$|G|$$, it follows that $$[x, G] = \langle C_{[x, G]}(a) \mid a \in Q \setminus \{1\} \rangle.$$ Therefore, $$G = C_G(x) \times [x, G] = \langle C_G(a) \mid a \in Q \setminus \{1\} \rangle.$$

• Lemma 2.2.2 Let $G$ be an abelian $p$-group and $A ≤ Aut(G)$ with $|A| = rq$, where $r$ and $q$ are prime numbers that may be equal. If $C_G(a) = 1$ for all $a ∈ A \setminus \{1\}$, then $A$ is cyclic.
– user1175180
Commented Apr 16 at 20:19
• Lemma 2.2.1 (b) Let $G$ be a $\pi$-group and $A$ a $\pi'$-group acting on $G$. (Suppose $G$ or $A$ is solvable.) Then, If $G$ is abelian, then $G=[G,A]\times C_G(A)$
– user1175180
Commented Apr 16 at 20:22

Since you have proved that $$H=[x, G]$$ is strictly smaller than $$G$$, you can now use your induction hypothesis to conclude that $$H= \langle C_H(a) \mid a \in Q \setminus \{ 1 \} \rangle$$. That is the first equation. The second equation doesn't rely on induction.

• Thanks for your quick answer! But I'm still confused over what the induction hypothesis is, since there is no base case. And why is it sufficent, that $,|H|$ is strictly smaller than $|G|$?
– user1175180
Commented Apr 16 at 21:21
• You can avoid the base case, by wording it as "Let $G$ be a group of smallest possible order such that the result is false". But in fact the result is vacuously true when $|G|=1$ because it is impossible to satisfy its hypotheses. Commented Apr 16 at 21:51
• Im sorry, but I still do not see, why we can say that $H= \langle C_H(a) | a \in Q \setminus \{1\} \rangle$, just because $|H| < |G|$. Do I have to repeat the steps of the proof for $[x,G]$ or something similar?
– user1175180
Commented Apr 16 at 22:45
• @Stippinator You don't need to repeat the steps of the proof. Your inductive hypothesis (using strong induction) is: Assume the statement is true for all groups strictly smaller than $G$. Since $H$ is a group that is strictly smaller than $G$ (and is necessarily an abelian $p$-group since it's a subgroup of an abelian $p$-group), the inductive hypothesis tells you without further work that the statement is true for $H$. Commented Apr 16 at 23:33
• @RobertShore thank you so much. I never heard of strong induction before. That makes it surprisingly easy.
– user1175180
Commented Apr 17 at 0:26