Let $G$ be a finite 2-group of nilpotency class two such that $\frac{G}{Z(G)}=\{Z(G), aZ(G), bZ(G), abZ(G)\}\simeq C_{2}\times C_{2}$. Then do there exist a non inner automorphism $\alpha$ of $G$ such that $\alpha(a)\neq a$, $\alpha(b)\neq b$ and $\alpha(ab)\neq ab$ ? For example this is true for $D_{8}$, dihedral group of order 8, or $Q_{8}$, generalized quaternion group of order 8.

  • $\begingroup$ The meaning of "acts trivial only on $Z(G)$" is really not clear. Why not write what you mean in mathematical language? Do you perhaps mean "Does there exists a non-inner automorphisms $\alpha$ of $G$ such that $\{g \in G \mid \alpha(g)=g \} = Z(G)$?" ? $\endgroup$ – Derek Holt May 29 '12 at 8:07
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    $\begingroup$ @DerekHolt: This is a follow-on from a comment I left to an earlier question of his (or, indeed, hers). $\endgroup$ – user1729 May 29 '12 at 9:12
  • $\begingroup$ Instead of re-posting your question, you could take it to the pro's? $\endgroup$ – user1729 Jun 6 '12 at 12:58
  • $\begingroup$ For future reference: instead of posting a new version of your question, you should instead edit the old version to improve it based on the comments posted. I've merged the old version into this one, since it appears you are providing more information to address @DerekHolt's comment. $\endgroup$ – Willie Wong Jun 6 '12 at 13:53
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    $\begingroup$ @user1729: I'd call Derek a pro. $\endgroup$ – j.p. Jun 7 '12 at 10:53

Yes there does exist such an automorphism. Note that the conditions $\alpha(a) \ne a$, $\alpha(b) \ne b$, $\alpha(ab) \ne ab$ imply that $\alpha$ is non-inner, because an inner automorphism $c_g$ must fix every element of $gZ(G)$.

If $Z(G)$ is not cyclic, then it contains a Klein 4-group $\langle x,y \rangle$ and we can define $\alpha(a)=ax$, $\alpha(b)=by$, $\alpha(ab)=abxy$, $\alpha(g)=g$ for all $g \in Z(G)$.

So suppose $Z(G) = \langle x \rangle$ with $|x|=n$. If $n=2$ we have $G=D_8$ or $Q_8$, which you know how to do.

If $|a| = |b|=2$, we can define $\alpha(a)=b$, $\alpha(b)=a$, $\alpha(x)=x$, so suppose that $|a|>2$.

If $|a|<2n$, then by replacing $a$ by $ax^i$ for suitable $i$, we get $|a|=2$. So suppose $|a|=2n$ and hence $\langle a \rangle$ is a cyclic subgroup of $G$ of index 2.

2-groups of order at least 16 with a cyclic subgroup of index two are known to be abelian, dihedral, semidihedral, generalized quaternion or modular, and the only one of these with $|G:Z(G)|=4$ is the modular group with presentation $\langle a,b \mid a^{2n}=b^2=1, a^b = a^{n+1} \rangle$. For this group, we can define $\alpha(a) = ab$, $\alpha(b) = ba^n$.


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