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

equivalently,

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.

D. N.
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Samuel Handwich
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  • http://igm.univ-mlv.fr/~fpsac/FPSAC07/SITE07/PDF-Proceedings/Posters/18.pdf –  Sep 24 '13 at 19:36
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    Extraspecial groups of order $p^3$ admit no fixed point free automorphisms. – Derek Holt Sep 24 '13 at 22:04
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    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. –  Sep 24 '13 at 22:18
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    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. –  Sep 25 '13 at 16:08
  • @SteveD: I double checked. there seem to be a ton. ForAny(a,f->ForAll(g,x->IsOne(x) or x^f <> x )); – Jack Schmidt Sep 25 '13 at 19:02
  • It's exam week, so I could be wrong, but I kind of remember there being a ton. Like x->x^2, y->y^2, z->z^4 has order 4. The only fixed point is 1. – Jack Schmidt Sep 25 '13 at 19:04
  • @JackSchmidt: oh yes, I made a stupid mistake, was looking at the group generated by the order-4 element to check for fpf. Just found 1550 order-4 fpf automorphisms, so you're spot on. –  Sep 25 '13 at 19:54
  • And in fact I see I have made a mistake: there are over 100 order-8 fpf automorphisms of that group of size 27. –  Sep 25 '13 at 20:00

1 Answers1

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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.

Samuel Handwich
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Yassine Guerboussa
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