If $G$ is a group and $X, Y \subseteq G$ then the commutator subgroup of $G$ is defined as $[G, G] = \langle [x, y] \mid x, y\in G \rangle$, where $[x, y] = x^{-1}y^{-1}xy$ and the group generated by commutator elements from $X$ and $Y$ is $[X, Y] = \langle [x, y] \mid x\in X, y\in Y\rangle$.

Now I am reading an article where the author talks about $[\alpha, A]$ where $\alpha$ is an automorphism of the group $A$. What does this mean? Is it related to the commutator group? The author does not define it, so I guess it is widely known what it is supposed to mean, but I haven't seen it before and I haven't had any luck looking in my books or Googling.


1 Answer 1


$[\alpha,A]$ means the subgroup $\langle \alpha(a^{-1})a \mid a \in A \rangle$ of $A$.

The connection with the commutator subgroups is that $[\alpha,A]$ is equal to the subgroup $[\langle \alpha \rangle, A]$ of the semidirect $A \rtimes \langle \alpha \rangle$ because, in the semidirect product $\alpha^{-1} a \alpha$ is equal to $\alpha(a)$.

  • $\begingroup$ Is there a typo here, $\alpha(a^{-1})a$ is just $\alpha$...? $\endgroup$
    – hayd
    Commented Mar 22, 2014 at 20:13
  • 1
    $\begingroup$ No, it is the product of $\alpha(a^{-1})$ and $a$. $\endgroup$
    – user129954
    Commented Mar 22, 2014 at 21:04

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