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Consider, for a group $G$, the subset $F$ consisting of all elements of $G$ that are fixed points of every possible automorphism of $G$. (Is there a term for such an element? I'd thought of "fixed element" but haven't so far found any evidence of the phrase being used with this meaning.)

One thing I have worked out is that $F$ will always be a subgroup of $G$. But there are two things that I've observed about all groups of order up to 15 but can't seem to prove or disprove in the general case:

  • $F$ is of order 1 or 2.
  • Every element of $F$ is of order 1 or 2.

Are these statements true? Obviously the first statement implies the second, but it might be the case that only the second is true.

If so, what is the proof?

If not, what counter-examples are there?

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    $\begingroup$ There is a group $G$ of order 16 with $F$ of order 4 and exponent 2. There is a group of order 63 with $F$ of order 3. There is a group of order 80 with $F$ cyclic of order 4. $\endgroup$ Commented Feb 23, 2014 at 16:55
  • $\begingroup$ @Jack Does cross-producting the groups of order 63 and 80 cross-product these subgroups also? That is, are there any additional automotphisms sneaking in or is it nice? (I am assuming you are using a computer, so hopefully this question isn't too unreasonable...) $\endgroup$
    – user1729
    Commented Feb 23, 2014 at 17:16
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    $\begingroup$ @user1729: in that particular case there are no extra automorphisms and F is cyclic of order 12. One has to be a little careful in general, for instance if G=C_2 then F=C_2, but GxG as F=1x1 not 2x2. $\endgroup$ Commented Feb 23, 2014 at 17:20
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    $\begingroup$ $F$ must be abelian. For every positive integer $n$ and non-negative integer $k$, there is a finite group group $G_{n,k}$ with $F=C_n \times (C_2)^k$. I'm having trouble with more general direct products, and haven't considered the infinite case. $\endgroup$ Commented Feb 23, 2014 at 20:37

2 Answers 2

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This subgroup has to be central, because of inner automorphisms. Now, when you quotient a non-abelian group by its centre then you must get a non-cyclic group (this is a common undergraduate exercise). Therefore, joining the dots, to prove your theorem you only need to consider the abelian groups. I will leave you to do this (the hint would be "consider the automorphism which inverts every element").

Edit Oh, sorry, you also need to look carefully at groups $G$ of order 12 with center of order 3 such that $G/Z(G)$ is the Klein 4-group.

Edit 2 Sorry, I have just realised that i mis-interpreted your question. I thought you wanted to prove thus for all groups of order less than 15. However, you can interpret my answer as saying "groups of order 15 have so few of these elements and one reason is because numbers less than 15 have so few factors." Thus the 12 issue. Try looking at, say, groups of order 18=2•3•3, or $2^23^25^2$. Or try using Gap to experiment some more.

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    $\begingroup$ What groups are these? The only non-abelian groups of order 12 are $D_6$ and $Dic_3$, whose centres have order 2, and $A_4$, which has centre of order 1 (by my calculation). $\endgroup$
    – Stewart
    Commented Feb 23, 2014 at 16:16
  • $\begingroup$ @Stewart I was unsure such a group existed, just that if it did it would be a possible conter-example that would need to be checked. $\endgroup$
    – user1729
    Commented Feb 23, 2014 at 17:02
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Another thing to observe that F is inside the center of G. (Since all automorphisms include the inner automorphisms)

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