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2:54 PM
6
Q: Are all algebraic automorphisms tame?

Yanior WegSuppose $G$ is a group. Lets call $\phi \in Aut(G)$ word automorphism iff $\exists n \in \mathbb{N} \{a_i\}_{i=0}^n \subset G \{e_i\}_{i=0}^n \subset \{-1; 1\} \forall t \in G \phi(t) = a_0t^{e_1}a_1…t^{e_n}a_n$. One can easily see, that all the word automorphisms form a normal subgroup in $Au...

 
 
4 hours later…
6:56 PM
For a semidirect product between Zpq and Zp, where Zpq is the normal subgroup and p > q, the commutator subgroup can be Zp or Zq or Zpq , right?
I mean we have to consider all 3 possibilities if no other condition is given right?
 

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