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I want to translate the following theorem from German to English:

5.12 Satz. Sei $\mathfrak{P}$ eine $p$-Gruppe und $\alpha$ ein Automorphismums von $\mathfrak{P}$ von su $p$ teilerfremdder Ordnung.*

  1. Für $p\gt 2$ lasse $\alpha$ alle Elemente der Ordnung $p$ von $\mathfrak{P}$ fest.
  2. Für $p\gt 2$ lasse $\alpha$ alle Elemente der Ordnungen $2$ und $4$ von $\mathfrak{P}$ fest.

Dann ist $\alpha =1$.

I used google translator and get the following:

Let B be a p-group and S be an automorphism of B of order coprime to p.

a) For p> 2, let S fix all the elements of B of order p.

b) for p = 2, let S fix all elements of orders 2 and 4 of B.

Then S = 1

I found the following translation on some research

For a finite $p$ group $P$, we write $$\Omega(P) = \left\{\begin{array}{ll} \Omega_1(P) &\text{if }p\gt 2\\ \Omega_2(P) &\text{if }p=2 \end{array}\right.$$ where $$\Omega_i(P) = \Bigl\langle x\in P \;\Bigm|\; |x| \bigm| p^i\Bigr\rangle.$$

Lemma 2.1 ([6] Kap. IV, Satz 5.12]). Suppose that a finite $p'$-group $A$ acts by automorphisms on a finite $p$-group $P$. If $A$ acts trivially on $\Omega_1(P)$ for $p\neq 2$ or on $\Omega_2(P)$ for $p=2$,then $A$ acts trivially on $P$.

I want to know if the last translation is true because I am not sure that if an automorphism comes from a $p^{'}$-group, then this automorphism is of order coprime to $p$.

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No, the theorem doesn't say an automorphism of a p-group has to be of order coprime to p. The theorem says that if its order is coprime to p and...etc. – DonAntonio Jun 30 '12 at 23:54
@DonAntonio: I Know what you mean. But, what about the Lemma 2.1, is it true? – user28083 Jul 1 '12 at 0:01
@user28083: If $A$ is a $p'$-group, then the elements are of order coprime to $p$. As automorphisms of $P$, they satisfy the hypothesis of the Theorem, so the conclusion follows: for each $\alpha\in A$, if $\alpha$ fixes all elements of order $p$, then it fixed everything (by the Theorem); so if this is true for all $\alpha\in A$, then it is true for all of $A$. For $p=2$, if each $\alpha\in A$ fixes all elements of order $2$ or $4$, then by the Theorem it must fix all of $P$, so the conclusion again follows. – Arturo Magidin Jul 1 '12 at 1:06
@ArturoMagidin: You mean that the automrphism induced by the elements of A are of orders coprime to p. – user28083 Jul 1 '12 at 1:44
@user28083: That follows from the fact that the elements of $A$ have order coprime to $p$. You have a homomorphism $A\to \mathrm{Aut}(P)$ (that's what "acts by automorphisms" means), and the image of $\alpha\in A$ has order that is a divisor of the order of $\alpha$. Hence, since the elements of $A$ have order coprime to $p$, so do their images in the automorphism group. – Arturo Magidin Jul 1 '12 at 2:13
up vote 2 down vote accepted

A better translation of the Theorem is:

Theorem 5.12. Let $P$ be a $p$-group, and let $\alpha$ be an automorphism of $P$ of order coprime to $p$.

  1. If $p\gt 2$ and $\alpha$ fixes all elements of order $p$; or
  2. If $p=2$ and $\alpha$ fixes all elements of orders $2$ or $4$;

then $\alpha$ is the identity.

The quoted lemma is equivalent to this result.

To see that the lemma follows from "Theorem 5.12", note that the fact that $A$ acts by automorphisms on $P$ is equivalent to saying that we have a group homomorphism $A\to \mathrm{Aut}(P)$. The conditions given on $A$ imply that if $\alpha\in A$, then the image of $\alpha$ in $\mathrm{Aut}(P)$ has order coprime to $p$ and satisfies the hypothesis of Lemma 5.12, hence $\alpha$ is the identity. Thus, the image of $A$ is trivial, so $A$ acts trivially on $P$.

Conversely, if the quoted Lemma is true, then Theorem 5.12 follows. Let $\alpha$ be an automorphism of $P$ of order coprime to $p$ that fixes each of the elements of order $p$ if $p\gt 2$, and the elements of orders $2$ and $4$ if $p=2$. Let $A=\langle \alpha\rangle$, viewed as a subgroup of $\mathrm{Aut}(P)$. Then $A$ acts on $P$ and satisfies the hypothesis of the lemma, hence acts trivially. In particular, $\alpha$ acts trivially; i.e., $\alpha$ is the identity automorphism. This proves "Theorem 5.12".

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