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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, $$G = C_G(x) \times [x, G] = \langle C_G(a) \mid a \in Q \setminus \{1\} \rangle.$$

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    $\begingroup$ 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. $\endgroup$
    – user1175180
    Commented Apr 16 at 20:19
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    $\begingroup$ 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 $G=[G,A]\times C_G(A)$ $\endgroup$
    – user1175180
    Commented Apr 16 at 20:22

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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.

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  • $\begingroup$ 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 $|G|$? $\endgroup$
    – user1175180
    Commented Apr 16 at 21:21
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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Apr 16 at 21:51
  • $\begingroup$ 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? $\endgroup$
    – user1175180
    Commented Apr 16 at 22:45
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    $\begingroup$ @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 $H$. $\endgroup$
    – Robert Shore
    Commented Apr 16 at 23:33
  • $\begingroup$ @RobertShore thank you so much. I never heard of strong induction before. That makes it surprisingly easy. $\endgroup$
    – user1175180
    Commented Apr 17 at 0:26

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