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Does every $p$-group of odd order admit fixed point free automorphisms?


Given an odd order $p$-group $P$, is there a group $C$ such that we can form a Frobenius group $P\rtimes C$?

Note that this is not true for $p$-groups of even order, for example $Q_8$ and $C_4$. But I cannot think of an example with odd order.

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Extraspecial groups of order $p^3$ admit no fixed point free automorphisms. – Derek Holt Sep 24 '13 at 22:04
To give an explicit example of Derek's comment: the extraspecial group of order $27$ (the one with all elements of order $3$) has a center of size $3$. Clearly the only prime $q$ which could act fpf on this center is $q=2$. But any group with an fpf automorphism of order $2$ is abelian, contradiction. To carry this further, if a $p$-group $G$ admits an fpf automorphism of prime order $q$, one can bound its nilpotency class in terms of $q$. Thus for any $p$, we can build $p$-groups of maximal class and large order which admit no fpf automorphism. – user641 Sep 24 '13 at 22:18
I think the extraspecial of order 125 and exponent 5 does have a fpf of order 4. – Jack Schmidt Sep 25 '13 at 15:42
@JackSchmidt: I just checked with GAP, I didn't find any. – user641 Sep 25 '13 at 16:08
up vote 5 down vote accepted

The following family of $p$-groups provides counter examples $$G = \langle a, b, c,d |a^{p^n}=b^{p^4}=c^{p^4}=d^{p^2}=1, [a,b]=[a,c]=b^{p^2}, [a,d]=c^{p^2}, [b,c]=a^{p^{n-2}}, [b,d]=c^{p^2}, [c,d]=c^{p^2} \rangle$$ with $p$ odd, and $n>3$.

For such a group $G$, every automorphism is central, that is $\operatorname{Aut}(G)$ acts trivially on $G/\operatorname{Z}(G)$. It is easy to see that the number of central automorphisms in that case (actually, for any finite group with no abelian direct factor) is equal to the order of $\operatorname{Hom}\left(G/G',\operatorname{Z}(G)\right)$. Thus $\operatorname{Aut}(G)$ is a $p$-group, so there is no automorphism acting fixed point freely on $G$.

The above example is due to V. Jain, P. Rai and M. Yadav.

As mentioned by Steve D, it is proved by U. Martin and G. Helleloid that (in some sense) almost finite $p$-groups have an automorphism group of $p$-power order, thus 'almost' $p$-groups have no fixed point free automorphisms.

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+1, excellent answer. – Andreas Caranti Sep 24 '13 at 21:20
Dear Professor Andreas, Thank you for your generous comment. – Yassine Guerboussa Sep 25 '13 at 9:30

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