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.
On automorphism of some finite 2-group of class nilpotency two
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0The 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 $\{g \in G \mid \alpha(g)=g \} = Z(G)$?" ? – 2012-05-29
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1@DerekHolt: This is a follow-on from a comment I left to an [earlier question](http://math.stackexchange.com/questions/150632/on-automorphisms-group-of-some-finite-2-groups#comment346988_150632) of his (or, indeed, hers). – 2012-05-29
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0Instead of re-posting your question, you could [take it to the pro's](http://mathoverflow.net/questions/ask)? – 2012-06-06
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0For 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. – 2012-06-06
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0This is almost different with earlier question. – 2012-06-07
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1@user1729: I'd call Derek a pro. – 2012-06-07
1 Answers
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 $\alpha(a) = ab$, $\alpha(b) = ba^n$.