Let $G$ be a finite 2-group such that $\mid Inn(G)\mid=4$, $Z(G)$ is not cyclic and $Z(G)$ has at least an element of order 4. Then prove or disprove there exists an $\alpha\in Aut(G)$ such that $\alpha(g)\neq g$ for some $g\in Z(G)$.

I think it is true.


It seems to me that such an automorphism always exists.

$G$ is minimal non-abelian, see e.g. Proposition 1.5 of this paper by Janko for the classification. (Added: this is not necessarily true, please stand by for a fix.)

Appealing to the classification, and since, by the assumption on the centre, the group $G$ is not quaternion of order 8, we have two cases,

  • $G = \langle a, b : a^{2^m} = b^{2^n} = 1, [a, b] = a^{2^{m-1}} \rangle$,
  • $G = \langle a, b, c : a^{2^m} = b^{2^n} = c^2 = 1, [a, b] = c, [a, c] = [b, c] = 1 \rangle$.

The centre of $G$ is $\langle a^2 \rangle \times \langle b^2 \rangle$ in the first case, and $\langle a^2 \rangle \times \langle b^2 \rangle \times \langle c \rangle$ in the second case.

Given the assumption on the exponent of $Z(G)$, in the second case we may assume without loss of generality that $m \ge 3$. Then $a \mapsto a^3, b \mapsto b$ defines an automorphism as requested.

In the first case, we have to distinguish two cases. If $m \ge 3$, then $a \mapsto a^3, b \mapsto b$ will do. If $n \ge 3$, then $a \mapsto a, b \mapsto b^3$ will do.

  • $\begingroup$ @ Andreas Caranti: I can not open link the paper by Janko. Please wirte name full of paper Janko. $\endgroup$ – maryam Jan 20 '13 at 18:22
  • $\begingroup$ @maryam Janko, Zvonimir. On minimal non-abelian subgroups in finite $p$-groups. J. Group Theory 12 (2009), no. 2, 289--303. $\endgroup$ – Andreas Caranti Jan 20 '13 at 18:26

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