Groups With No Automorphisms

We classify the groups with trivial automorphism groups. This was a pretty tricky problem on my girlfriend’s algebra qualifying exam the other day!

Proposition. If \(G\) is a group with trivial automorphism group, then \(G\) is isomorphic to either the identity group or \(\mathbb Z/2\mathbb Z\).

Proof. We proceed in steps.

1. We claim that \(G\) is abelian. Because \(\operatorname{Aut}(G)\) is trivial, we see that \(\operatorname{Inn}(G)\) is also trivial. We now claim that \(\operatorname{Inn}(G)\cong G/Z(G)\), from which the claim will follow because then \(G=Z(G)\) is an abelian group. Well, \(\operatorname{Inn}(G)\) is the image of the map \(\gamma_\bullet\colon G\to\operatorname{Aut}(G)\) given by \[\gamma_g\colon h\mapsto hgh^{-1}.\] Thus, \(\operatorname{Inn}(G)\cong G/Z(G)\) follows because the kernel of this map is \[\left\{g\in G:hgh^{-1}=g\text{ for all }h\in G\right\},\] which is the center.

2. We claim that every element of \(G\) has order dividing \(2\). Indeed, we will show that any abelian group \(G\) with an element of order bigger than \(2\) has a nontrivial automorphism. For this, let \(\iota\colon G\to G\) be the function \(\iota(g):= g^{-1}\). Because $G$ is abelian, we see that \(\iota(gh)=\iota(g)\iota(h)\), so one can check that \(\iota\) is in fact a homomorphism. Because \(\iota^2=\operatorname{id}_G\), we see that \(\iota\) is an automorphism. Lastly, to see that \(\iota\) is nontrivial, recall that we have an element \(g\in G\) of order larger than \(2\). Thus, \(g\ne g^{-1}\), so \(\iota(g)\ne g\), so \(\iota\ne\operatorname{id}_G\).

3. We show that \(G\) is isomorphic to the identity group or \(\mathbb Z/2\mathbb Z\). The previous two steps imply that every element of \(G\) has order dividing \(2\). Thus, \(G\) is an \(\mathbb F_2\)-module, so \(G\cong\mathbb F_2^{\oplus\alpha}\) for some ordinal \(\alpha\). We would like to show that \(\alpha\in\{0,1\}\) if \(\operatorname{Aut}(G)\) is trivial. Well, note that a function \(\varphi\colon G\to G\) is a group homomorphism if and only if it is \(\mathbb Z\)-linear, which is equivalent to being \(\mathbb F_2\)-linear in this case. It follows that \[\operatorname{Aut}(G)\cong\operatorname{GL}_\alpha(\mathbb F_2).\] We now see that the right-hand group is nontrivial whenever \(\alpha\ge2\): for example, there is a nontrivial \(\mathbb F_2\)-linear bijection given by switching the first two coordinates of our basis. It follows that \(\alpha\in\{0,1\}\).

The above steps complete the proof. \(\blacksquare\)

Remark. In fact, a more general version of the first step is true: if \(\operatorname{Inn}(G)\) is cyclic, then \(G\) is abelian. To see this, note the hypothesis implies that \(G/Z(G)\) is abelian. Now, suppose that \(G/Z(G)\) is generated by \(gZ(G)\) for some \(g\in G\). It will be enough to check that \(g\in Z(G)\). Well, any element in \(G\) can be written as \(g^nh\) for some \(n\in\mathbb Z\) and \(h\in Z(G)\), which does commute with \(g\). Thus, \(g\in Z(G)\) follows.