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