Some interesting problems in Group theory
There is an interesting problem.
Question. Can $\mathfrak{S}_n$ be isomorphic to $ [G,G]$ for some group $G$?
The answer is no for $n\geq 3$.
Any such $G$ induces a group homomorphism $G\to \operatorname{Aut}(\mathfrak{S}_n)$ by conjugation. We know that $\operatorname{Aut}(\mathfrak{S}_n)=\operatorname{Inn}(\mathfrak{S}_n)\cong \mathfrak{S}_n$ except $n=2,6$. When $n=6$, the outer automorphism is of order $2$, so
$[\operatorname{Aut}(\mathfrak{S}_n),\operatorname{Aut}(\mathfrak{S}_n)]\cong \mathfrak{A}_n$ (the case of $n=6$ due to the classification of group of order 4).But $\mathfrak{S}_n=[G,G]$ acts on itself with image $\mathfrak{S}_n$ a contradiction.
We can generalize the trick to prove this
Generalization. If $S$ is a nonabelian simple group with outer automorphism group of solvable length $n$. If $G$ with $G^{(n+k)}=S$ for some $k>0$, then $$G^{(n)}=G^{(n+k)}\oplus G^{(n)}/G^{(n+k)}. $$
The process is not hard.
Note that $(\operatorname{Aut}S)^{(n)}=\operatorname{Inn}S$ since $\operatorname{Inn}S\cong S$ is not solvable.
By the action of conjugation, we get a map $G\to \operatorname{Aut}(S)$. Then $G^{(n)}$ and $G^{(n+k)}$ has the same image $\operatorname{Inn}S$. Now we have $$\begin{array}{ccccc}S=& G^{(n+k)}& \to &\operatorname{Inn}S & =S \\ & \downarrow && \downarrow \\ & G^{(n)}&\to & \operatorname{Inn}S & =S\end{array}$$So $G^{(n)}=G^{(n+k)}\oplus G^{(n)}/G^{(n+k)}$.
Here are some remarks,
- By classification of finite simple groups (Schreier conjecture), $n\leq 4$.
-
For $\mathfrak{A}_n$, $\operatorname{Out} \mathfrak{A}_n$ is of order $2$, for $n\neq 1,2,6$, trivial for $n=1,2$, and of order $4$ for $n=6$. So the solvable length is $1$.
I raised a similar problem.
Problem. For $p\geq 3$, the group $$T_n(\mathbb{F}_p)=\left(\begin{matrix}\mathbb{F}_p^\times & \cdots & \mathbb{F}_p \\ & \ddots & \vdots \\ && \mathbb{F}_p^\times\end{matrix}\right)$$ of invertible upper triangle matrices over $\mathbb{F}_p$ is not isomorphic to $[G,G]$ for some group $G$.
If you could compute the automorphism of $T_n(\mathbb{F}_p)$, and show that its commutator does not contain all inner automorphism, it will follow by a similar argument as above. But this maybe only can be done by Démon de Laplace. We only need to find a contradiction.
Firstly, the group $$H=\left(\begin{matrix}1& 0 &\mathbb{F}_p \\ & \ddots & 0 \\ & & 1\end{matrix}\right)\cong \mathbb{F}_p$$ is characteristic (stable under automorphism), since it is the center of Sylow $p$-subgroup. Now any such $G$ induces $G\to \operatorname{Aut}(\mathbb{F}_p)=\mathbb{F}_p^\times$. Then $[G,G]$ acts trivially on $H$. But when $x\in \mathbb{F}_p^\times\setminus \{1\}$, the conjugation by $\operatorname{diag}(x,1,\ldots,1)$ acts as $x\in \mathbb{F}_p^\times$.
About this idea, I have three remarks,
- The first question was initially asking for $\mathfrak{S}_4$, where I took the Klein 4 group as characteric subgroup. I want to ``use the trick of characteristic subgroup'' so I raised this problem.
- I though the group $H$ in the proof is the $(n-1)$-th commutator, but not generally true, for example $T_2(\mathbb{F}_2)$ which is abelian. Liu Ben told this problem.
- The same argument holds also for $T_n(\mathbb{Q})$, dihedral groups $D_{2n}$. They both have a cyclic characteristic subgroup with inner automorphism acting nontrivially.
Here are more ``continous'' remarks that the above problems reflect some interesting results.
- For a finite dimensional complex Lie algebra $\mathfrak{g}$, if it is solvable, then $[\mathfrak{g},\mathfrak{g}]$ is nilpotent by Cartan criteria. In other words, if $\mathfrak{g}$ is solvable but not nilpotent, then it is not a Lie bracket of any finite dimensional complex Lie algebra.
- Similarly, the similar result holds for algebraic groups. An algebraic group $G$ over algebraic closed field, is solvable, then $[G,G]$ is unipotent (then nilpotent).
The argument above seems to be an analogy of this for abstract group. For abstract group, one may dream that the derived group of solvable group is nilpotent, but this centainly cannot be true.
Taking this into accout, we can view the first problem from another point of view. In some sense, $\mathfrak{S}_n$ is parallel to $\operatorname{GL}_n$, and $\mathfrak{A}_n$ to $\operatorname{SL}_n$.
- For algebraic groups, when $G$ is reductibe, then $[G,G]$ is semisimple. We know that if $[G,G]$ is solvable, then so is $G$. In all, any algebra group which is reductive but not semisimple is not a commutator of an algebraic group.
So the first argument is an analogy of this for field of one element.
Last, maybe less relative, the dual question.
Exercise. Let $Q_8$ be the group of units of Cayley algebra $\{\pm 1, \pm i, \pm j,\pm k\}$. Then $Q_8$ is not of the form $G/Z(G)$.
Remind a classic result, $G/Z(G)$ cannot be cyclic. Because one can find if so, the center is bigger than $Z(G)$.
Because, if we take some preimage of $i$, say $x$, and any preimage of $-1$, say $y$, we will have $x^2c=y$ for some $c\in Z(G)$. Now $xy=x^3c=yx$. So it is easy to see $y$ actually commutes with all elements in $G$, a contradiction.
Similarly, one can prove that $\mathbb{Q}$ is not of the form $G/Z(G)$.
If $\mathbb{Q}=G/Z(G)$. Pick the preimage of $1$, say $x$, and preimage of $1/q$, say $y$. Then $y^qc=x$ for some $c\in Z(G)$, so $xy=yx$. So $xy^p=y^px$. So we have a set of representative elements for $G/Z(G)$ commutes with $x$, so $x\in Z(G)$, a contradiction.
