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Well, this is an exercise problem from Herstein which sounds difficult:

  • How does one prove that if $|G|>2$, then $G$ has non-trivial automorphism?

The only thing I know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism group of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?

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    $\begingroup$ You've got it switched. The group of inner automorphisms is isomorphic to G/Z(G). This essentially solves the problem. $\endgroup$
    – Qiaochu Yuan
    Oct 30, 2010 at 19:21
  • $\begingroup$ @Qiaochu Yuan: Sorry $\endgroup$
    – anonymous
    Oct 30, 2010 at 19:23
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    $\begingroup$ So figure it out. Inner automorphisms are automorphisms, so if G/Z(G) is nontrivial there is an inner automorphism. Otherwise... $\endgroup$
    – Qiaochu Yuan
    Oct 30, 2010 at 19:25
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    $\begingroup$ @Chandru1: So now you have reduced to the case when $G$ is abelian. Can you think of an automorphism that works for abelian groups, but not for nonabelian groups? $\endgroup$
    – Jonas Meyer
    Oct 30, 2010 at 19:27
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    $\begingroup$ @Chandru1: Please think more about where it is reduced to the abelian case. Yes, that is the map I had in mind. $\endgroup$
    – Jonas Meyer
    Oct 30, 2010 at 19:29

3 Answers 3

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As you note in the question, the group of inner automorphisms Inn($G$) is isomorphic to $G/Z(G)$. In particular, it's trivial if and only if $Z(G)=G$. So there is a non-trivial (inner) automorphism unless $G=Z(G)$.

Now, notice that, by definition, $Z(G)=G$ if and only if $G$ is abelian; so we have reduced to the abelian case.

If $G$ is abelian then $g\mapsto -g$ is an automorphism, and it is non-trivial unless $g=-g$ for all $g\in G$. But $g=-g$ if and only if the order of $g$ divdes two. So we have now reduced to the case in which $2g=0$ for all $g\in G$.

In this case, $G$ is a vector space over the field $\mathbb{Z}/2$. As $|G|$ is equal to 2 raised to the power of the $\mathbb{Z}/2$-dimension of $G$, the hypothesis that $|G|>2$ implies that $\mathrm{dim}_{\mathbb{Z/2}} G>1$. But now we can write down lots of linear automorphisms of $G$. For instance, you could fix any basis $g_1,g_2,\ldots$ and take the automorphism $g_1\mapsto g_2$, $g_2\mapsto g_1$ and $g_i\mapsto g_i$ for every $i>2$.

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    $\begingroup$ HJRW, I'm sorry, but I couldn't grasp the details of your reasoning when $G$ is abelian and $g^2 = e$ for all $g\in G$. I've never encountered this vector space business getting into group theory. So can you please explain your reasoning to me in more elementary terms? $\endgroup$
    – Saaqib Mahmood
    Jun 14, 2014 at 13:59
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    $\begingroup$ @SaaqibMahmuud; well, suppose that $G$ is abelian and $2g=0$ for all $g$. Then you can check very easily that your group $G$ satisfies the axioms of a vector space over the field with two elements (which I denoted by $\mathbb{Z}/2$ in my answer). Then you argue exactly as I described above. If there are any further steps you have difficulty with, it would help if you said which ones they are! $\endgroup$
    – HJRW
    Jun 15, 2014 at 5:46
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    $\begingroup$ What is this fundamental theorem of abelian groups? I know a fundamental theorem of finitely generated abelian groups. In the infinitely generated case I think you just want to point out that any two distinct elements generate a Klein 4-group, which has automorphisms and is a term in a direct product decomposition. Actually, I'm not sure the direct product decomposition is clear here. $\endgroup$
    – Kevin Arlin
    Oct 17, 2014 at 3:03
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    $\begingroup$ @tmastny, the $g_i$ are a basis for $G$, not an enumeration of $G$. So, by definition, one computes the action of $T$ on an element $g\in G$ by writing $g$ as a product of $g_i$'s and seeing what happens. In particular, in your example, $T(g_1g_i)=T(g_1)T(g_i)=g_2g_i$. $\endgroup$
    – HJRW
    Jan 16, 2015 at 10:03
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    $\begingroup$ @KevinCarlson, it might not be a direct product. For instance, it might be a direct sum... $\endgroup$
    – HJRW
    Jan 16, 2015 at 11:06
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If $G$ is not abelian, then conjugation by a noncentral element will do.

If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism. It will be nontrivial unless every element of $G$ equals its inverse, that is, if every element of $G$ is of exponent $2$.

If every element of $G$ is of exponent $2$, then $G$ is a vector space over the field of $2$ elements, so it is isomorphic to a (possibly infinite) sum of copies of $C_2$, the cyclic group of two elements. Since $|G|\gt 2$, there are at least two copies, so the linear transformation that swaps two copies of $C_2$ is a nontrivial automorphism.

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    $\begingroup$ Beat me to it by 90 seconds! $\endgroup$
    – HJRW
    Oct 30, 2010 at 21:17
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    $\begingroup$ Any idea whether this can be proven constructively? $\endgroup$
    – Carl Mummert
    Oct 30, 2010 at 23:49
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    $\begingroup$ Carl, what do you mean? There are three cases, but in each case the automorphism is completely explicit. The existence of complete groups, in which every automorphism is inner, shows that there is no one construction that can work for all groups (as abelian groups have no inner automorphisms). $\endgroup$
    – HJRW
    Oct 31, 2010 at 1:05
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    $\begingroup$ @Arturo Magidin: in your comment of 2010-10-31 01:59:58Z, the set $H$ is not a subgroup of $G$, right? I don't see why the map you define has to be an automorphism. $\endgroup$
    – Carl Mummert
    Oct 31, 2010 at 21:50
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    $\begingroup$ @Arturo Magidin: Sorry; it's hard to read the TeX and it isn't rendering for some reason. Suppose that $a$ is not in $K = \langle x,y\rangle$. Then $a + x$ is also not in $K$, because $x$ is in $K$. So the subgroup generated by the elements not in $K$ contains $x$. In general we would have to make $H$ a complementary subgroup to $K$, which does exist in the setting at hand, but I'm not certain that there has to be a complementary subgroup that is computable if $G$ is computable. $\endgroup$
    – Carl Mummert
    Oct 31, 2010 at 22:28
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The other two answers assume the axiom of choice:

  • Arturo Magidin uses choice when he forms the direct sum ("...it is isomorphic to a (possibly infinite) sum of copies of $C_2$...")
  • HJRW uses choice when he fixes a basis (the proof that every vector space has a basis requires the axiom of choice).

If we do not assume the axiom of choice then it is consistent that there exists a group $G$ of order greater than two such that $\operatorname{Aut}(G)$ is trivial. This is explained in this answer of Asaf Karagila.

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    $\begingroup$ Asak? Who's that guy? :) $\endgroup$
    – Asaf Karagila ♦
    Aug 30, 2018 at 10:32
  • $\begingroup$ Gah! I typed your first name by hand, then copied-and-pasted your second name... incidentally, I noticed that you've started/about to start a postdoc at UEA. It a nice place - I was at a workshop there earlier this year. There is a theatre on the campus which serves nice brownies. $\endgroup$
    – user1729
    Aug 30, 2018 at 10:42
  • $\begingroup$ Yeah, I like Norwich so far. $\endgroup$
    – Asaf Karagila ♦
    Aug 30, 2018 at 10:45

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