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Let $G$ be finite group and $Aut(G)$ denote the automorphisms group of $G$.

1) If $\mid G\mid>2$ is even, then do $\mid Aut(G)\mid$ is even?

2) Do we can determine structure all finite groups such that the order of the automorphisms group is even?

I want the papers about this subject. Please Help me

maryam
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    If $C_2$ is the cyclic group of order two then $\operatorname{Aut}(C_2)$ is trivial, and so has odd order. – user1729 Apr 10 '13 at 15:59
  • Ok. I forgot to write that $\mid G\mid >2$. I must edit it. – maryam Apr 10 '13 at 16:02
  • Do we need the condition $|G|$ is even? (Interestingly, GAP says we do. Without this condition, the smallest non-trivial counterexample is of order $3^6=729$, and there's many of this order.) – Douglas S. Stones Apr 10 '13 at 16:03
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    More seriously, $G/Z(G)$ is a subgroup of $\operatorname{Aut}(G)$. Thus, if $G/Z(G)$ is even then so is $\operatorname{Aut}(G)$. For example, if $G$ is a non-abelian group of order $2^n$ then $\operatorname{Aut}(G)$ is even. – user1729 Apr 10 '13 at 16:03
  • @DouglasS.Stones: No. Take $C_3\times C_3$. – user1729 Apr 10 '13 at 16:04
  • In first question $G$ is even. But in later question G can be odd. – maryam Apr 10 '13 at 16:06
  • Oh, also, any abelian group has even order automorphism group (or is cyclic of order two)! This is because inverting every element is an automorphism. This is the trivial automorphism if only if your group is $C_2\times C_2\times\ldots C_2$, and then you can swap two of the factors, unless there is only one factor. If there is only one factor then your group is cyclic of order two. Done. – user1729 Apr 10 '13 at 16:07
  • @user1729: Do we can to say if $\mid G\mid$ is even, then $\mid \frac{G}{Z(G)}\mid$ is even? – maryam Apr 10 '13 at 16:09
  • @maryam: I do not know. The Feit-Thompson theorem is relevant though. It is a famous result, which states that every group of odd order is soluble, and so has non-trivial centre. If there was a group $H$ of odd order and trivial centre then you could form $G:=H\times C_2$ and then $G/Z(G)$ would have odd order even though $G$ has even order. But this doesn't work. By Feit-Thompson. – user1729 Apr 10 '13 at 16:14
  • @user1729: there is a group of order 21 with trivial centre. – Derek Holt Apr 10 '13 at 16:23
  • @DerekHolt: Seriously? Why? (Oh...am I getting nilpotent confused with soluble?...soluble doesn't imply non-trivial centre, does it?) – user1729 Apr 10 '13 at 16:23

1 Answers1

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The answer to 1) is no, but it is not easy to find examples. In the paper

Dark, R. S. A complete group of odd order. Math. Proc. Cambridge Philos. Soc. 77 (1975), 21–28.

a group $H$ of order $3 \times 19 \times 7^{12}$ is constructed, which has trivial centre and is isomorphic to its own automorphism group. So $H \times C_2$ has even order but with odd order automorphism group.

Note that it $|G|$ is even and $|G/Z(G)|$ is odd, then $G$ is a direct product of a group of odd order and an abelian 2-group $T$, and if $|{\rm Aut}(G)|$ is odd then we must have $|T|=2$. So all examples must be similar this one (although the odd order direct factor would not necessarily have to be complete).

As for 2), that sounds as hard as determining the structure of all finite groups!

Derek Holt
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    The smallest examples of groups with nontrivial automorphism groups of odd order are of order $3^6$. For example, $G = \operatorname{SmallGroup}(3^6, 90)$ has an automorphism group of order $3^7$. So by your remark, $C_2 \times G$ is a smallest possible example of even order. Reference: D. MacHale and R. Sheehy, Finite groups with odd order automorphism groups, Math. Proc. R. Ir. Acad. 95A (1995), no. 2, 113–116. JSTOR – Mikko Korhonen Apr 10 '13 at 17:19