I recently attended an interview for admission to graduate programs in Mathematics. The interviewing professor asked me a question - Tell me a group whose group of automorphisms is non-abelian.

Because I was too nervous, I couldn't think of anything substantial.

Could someone please tell me, in such a situation, what is the logical way of deducing the answer to such a question posed by the interviewer?

Thank you very much for your help!

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    $\begingroup$ Any complete group, i.e., a group such that all automorphisms are inner and the center is trivial so that $G\rightarrow Aut(G)$ mapping each $g$ to the conjugation by $g$ map is an isomorphism. $\endgroup$ – Taylor Jul 2 '15 at 13:40
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    $\begingroup$ @Taylor Trivial centre (and non-trivial $G$) suffices, since then $\operatorname{Aut} G$ contains a subgroup that is isomorphic to $G$. $\endgroup$ – Daniel Fischer Jul 2 '15 at 13:42

The littlest counter-example :

$$Aut(\frac{\mathbb{Z}}{2\mathbb{Z}}\times \frac{\mathbb{Z}}{2\mathbb{Z}})=S_3 $$

All three non-zero elements are freely permuted.

Now, with such a question, you have two ways, either you take a non-commutative group with trivial center, in which case you will have $G=G/Z(G)=Inn(G)$ hence a non commutative subgroup of $Aut(G)$ or you can also consider the following property which is not hard to show: A $\underline{\text{finite}}$ Abelian group $A$ has an abelian automorphism group if and only if $A$ is cyclic.

Something less intuitive : There exists non-abelian groups with abelian automorphism group.

The proof of $A$ abelian non cyclic implies that $Aut(A)$ is non commutative goes as follows : we know that an abelian group $A$ is the direct product of its $p$-Sylows :

$$A=S_1\times...\times S_r$$

Clearly (it should be checked, the $p$-Sylows are characteristic) this leads to :

$$Aut(A)=Aut(S_1)\times ...\times Aut(S_r)$$

Hence, it suffices to check the property for abelian $p$-groups. Now assume $A$ is an abelian $p$-group and that $A$ is non-cyclic then we have that :

$$A=\frac{\mathbb{Z}}{p^{a_1}}\times...\times \frac{\mathbb{Z}}{p^{a_r}}$$

and $r\geq 2$. Now define :

$$B_1=\frac{\mathbb{Z}}{p^{a_1}}\times...\times \frac{\mathbb{Z}}{p^{a_{r-2}}}$$

$$B_2=\frac{\mathbb{Z}}{p^{a_{r-1}}}\times \frac{\mathbb{Z}}{p^{a_r}}$$

We have that $Aut(A)$ contains $Aut(B_1)\times Aut(B_2)$. I will show that $Aut(B_2)$ is non commutative. Define $e_1$ and $e_2$ to be respectively the element $(1,0)$ and $(0,1)$ of respective orders $p^{a_{r-1}}$ and $p^{a_r}$. Now you have two automorphisms :

$$\phi_1\text{ defined by }\phi_1(e_1):=e_1\text{ and } \phi_1(e_2):=e_1+e_2 $$

$$\phi_2\text{ defined by }\phi_2(e_1):=e_1+p^{a_r-a_{r-1}}e_2\text{ and } \phi_1(e_2):=e_2 $$

I claim that this defines group morphisms (it suffices to check that $\phi_i(e_1)$ is of order $p^{a_{r-1}}$ and $\phi_i(e_2)$ is of order $p^{a_{r}}$). Secondly they are surjective hence bijective so they are group automorphisms of $B_2$. Now :

$$\phi_1\circ\phi_2(e_2)=e_1+e_2 $$

$$\phi_2\circ\phi_1(e_2)=\phi_2(e_1+e_2)=e_1+(1+p^{a_r-a_{r-1}})e_2 $$

Clearly we have : $\phi_1\circ\phi_2\neq \phi_2\circ\phi_1$. Hence we defined two non-commuting automorphisms which imply that $Aut(B_2)$ is non commutative hence $Aut(A)$ is non commutative as well.

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    $\begingroup$ A link on the less intuitive fact you mention: MathOverflow: when is Aut(G) abelian $\endgroup$ – Jeppe Stig Nielsen Jul 3 '15 at 13:21
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    $\begingroup$ "or you can also consider the following property which is not hard to show: An Abelian group $A$ has an abelian automorphism group if and only if $A$ is cyclic." That's false, the additive group of the rational numbers is a counterexample. In fact, even $A$ abelian and $\operatorname{Aut} A$ cyclic does not imply that $A$ is cyclic. (Of course you can fix your answer by adding one word...) $\endgroup$ – Pete L. Clark Jul 3 '15 at 16:15
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    $\begingroup$ @PeteLClark, you are very right. Thanks for correcting it. I was really thinking finite groups but it was implicite. $\endgroup$ – Clément Guérin Jul 3 '15 at 16:56
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    $\begingroup$ See my answer for a generalization. $\endgroup$ – Martin Brandenburg Jul 3 '15 at 22:40
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    $\begingroup$ @ClémentGuérin Could you please give me a hint to show that if $A$ is a finite Abelian group with abelian automorphism group, then $A$ is cyclic? $\endgroup$ – Yilong Zhang Jul 9 '15 at 3:29

Let $A$ be any nontrivial group. Then $\operatorname{Aut}(A\times A\times A)$ certainly has $S_3$ as a subgroup (per permutation of the summands)

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    $\begingroup$ Of course the real question was not to name such a group, but to describe how to find such a group. I find it natural to try the simplest example of a nonabelian group ($S_3$) and try to ensure that $S_3$ is a subgroup ioof the automorphism group. As permuting things is the favorite pastime of $S_3$, something like above might jump to your mind. $\endgroup$ – Hagen von Eitzen Jul 2 '15 at 13:46
  • $\begingroup$ Yeah, I get it. I now wish it had striked me there as well. :) $\endgroup$ – MathMan Jul 2 '15 at 16:51
  • $\begingroup$ @HagenvonEitzen : So in general , $\ Aut (G^n)$ is not abelian for any $n>2$ , for any non-trivial group $G$ , right ? $\endgroup$ – user228168 May 7 '16 at 10:01

View a vector space as an abelian group. Then the general linear group will be contained in the group of automorphisms of the vector space. Since the general linear group is usually not abelian, this gives a lot of natural examples.

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    $\begingroup$ Thank you for the answer :) $\endgroup$ – MathMan Jul 2 '15 at 16:50
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    $\begingroup$ This is the most sensible answer. $\endgroup$ – goblin Jul 2 '15 at 17:26
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    $\begingroup$ $S_{3} \cong {\rm GL}(2,2)$ is actually the smallest case. $\endgroup$ – Geoff Robinson Jul 10 '15 at 10:12

Perhaps the easiest example is the group $S_3$, which has only inner automorphisms and has trivial center. We have $$ Aut(S_3)=Inn(S_3)\simeq S_3, $$ which is nonabelian.

  • $\begingroup$ Thanks for your answer. Is there a theorem which states the isomorphism between $Inn (S_3)$ and $S_3$? Apologize if it's too trivial. $\endgroup$ – MathMan Jul 2 '15 at 13:45
  • $\begingroup$ I think it's because $|Inn S_3| = 6$ and any group whose order is of the form $2p$ is isomorphic to either $Z_{2p}$ or $D_p$ . Am I correct? $\endgroup$ – MathMan Jul 2 '15 at 13:46
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    $\begingroup$ Yes, that is correct. If you need another example, then take the quaternion group $Q_8$. We have $Aut(Q_8)\simeq S_4$. $\endgroup$ – Dietrich Burde Jul 2 '15 at 13:47
  • $\begingroup$ Got it. Thank you for answering :) $\endgroup$ – MathMan Jul 2 '15 at 16:51
  • $\begingroup$ In general, $Inn(G) \cong G/Z(G)$. But $S_n$ has trivial center for $n \geq 3$. So $Inn(S_3) \cong S_3$ is trivial. The equation $Aut(S_3)=Inn(S_3)$ needs some thought, though. It is quite interesting that $Aut(S_n)=Inn(S_n)$ holds for all $n \neq 6$, but not for $n=6$. $\endgroup$ – Martin Brandenburg Jul 10 '15 at 10:42

If $G$ is any non-trivial group, then $\mathrm{Aut}(G \times G)$ is not abelian.

This is an improvement of Hagen von Eitzen's observation.

Proof. If $G$ is abelian, then $(x,y) \mapsto (y,x)$ does not commute with $(x,y) \mapsto (x,x+ y)$.

If $G$ is not abelian, we find a nontrivial inner automorphism $\sigma$ of $G$. But then $(x,y) \mapsto (y,x)$ does not commute with $(x,y) \mapsto (x,\sigma(y))$.


$Aut(C_p \times C_p)$ coincides with the set of invertible linear transformations and so is $GL(2,\mathbb F_p)$, which is not abelian.

  • $\begingroup$ See my answer for a generalization. $\endgroup$ – Martin Brandenburg Jul 3 '15 at 22:40

If they are talking about the morphisms of a group, here is a approach.

Take automorphism group of the free group of the set $\mathbb Z$. It's automorphism group is a supergroup of the permutation group of $\mathbb Z$.

This method is very general, working for rings, monoids, and abelian groups as well. A function can always be made into a homomorphism between free groups, (or other structures.) This is based in categorical thinking. Anytime they ask about morphisms, think of any category theory you know.

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    $\begingroup$ Thank you for your answer :) $\endgroup$ – MathMan Jul 2 '15 at 16:49
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    $\begingroup$ The best part is you don't really need to know any group theory. This works for structures besides groups. $\endgroup$ – PyRulez Jul 2 '15 at 17:31
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    $\begingroup$ "Take automorphism group of the free group of the set $\mathbb{Z}$. Its automorphisms correspond exactly to the permutations of $\mathbb{Z}$." That's not true. Any permutation of $S$ induces an automorphism of $\operatorname{Free}(S)$, but this is not the entire automorphism group. Your statement is analogous to the (perhaps more evidently false) statement that the group of automorphisms of an $n$-dimensional vector space over a field corresponds to the permutations of a given basis. $\endgroup$ – Pete L. Clark Jul 3 '15 at 16:13
  • $\begingroup$ Pate's right. Still, this is a great approach which answers the question, since the automorphism group of the free group will contain a copy of the symmetric group (and the homomorphisms of the free group on $\mathbb{Z}$ will contain the collection of functions on $\mathbb{Z}$). $\endgroup$ – Seth Jul 3 '15 at 22:05
  • $\begingroup$ @PeteL.Clark Thanks, Fixed. $\endgroup$ – PyRulez Jul 3 '15 at 22:12

Well, for instance, the maps $$\varphi_a:G\to G\\ \varphi_a(x)=a^{-1}xa$$ are (inner) automorphism.

Now, let $a,b$ such that $\varphi_a\varphi_b=\varphi_b\varphi_a$. We have $$\begin{cases}\varphi_a\varphi_b(b)=a^{-1}b^{-1}bba=a^{-1}ba\\\varphi_b\varphi_a(b)=b^{-1}a^{-1}bab\end{cases}$$ And so it must hold

$$a^{-1}ba=b^{-1}a^{-1}bab\implies b(a^{-1}ba)=(a^{-1}ba)b$$

And, I mean, in a general non-abelian group you cannot expect all the elements to commute with all their conjugates. Counterexaples arise naturally, for instance, in linear groups over most fields, due to the following lemma:

Two diagonalisable matrices commute if and only if they are diagonalisable simultaneously.

Explicitly, you can pick $G=\operatorname{GL}(2,\mathbb F_3)$, with $b=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\ a^{-1}ba=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and use the lemma, yielding that $\operatorname{Inn}(\operatorname{GL}(2,\mathbb F_3))$ is not abelian.

  • $\begingroup$ I get it. :) Thank you for the answer. $\endgroup$ – MathMan Jul 2 '15 at 16:50

$\DeclareMathOperator{Aut}{\operatorname{Aut}}$$\Aut(\Bbb C,+)$ contains the conjugation mapping $z \mapsto z^*$ and the multiplication by $i$ mapping as automorphisms. Now notice that $(iz)^* = -iz^* \neq iz^*$ for $z \neq 0$.

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    $\begingroup$ I hope that you are OK with my edit. If not, feel free to unmake it. $\endgroup$ – Martin Brandenburg Jul 10 '15 at 10:05

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