# Automorphism preserving a subgroup [closed]

Let $$G$$ be a group with a subgroup $$H$$ such that the index of $$H$$ is $$n$$. For any automorphism $$\tau$$ of $$G$$, can we prove that $$\tau^{n!}$$ preserves $$H$$? Here $$\tau^{n!}$$ means the self-composition of $$\tau$$ for $$n!$$ times.

$$\tau$$ acts on the set of subgroups of index $$n$$. There can be many such subgroups, so we cannot expect something like you suggest. Indeed, let $$G = (\Bbb Z/n\Bbb Z)^M$$ and let $$H$$ be the kernel of projection to the first factor, and let $$\tau$$ be cyclic permutation of the $$M$$ components. Then the order of $$\tau$$ in $$\operatorname{Aut}(G)$$ is $$M$$ (and for no smaller power is $$H$$ preserved), which may be $$>n!$$.
The smallest counterexample is the following: let $$G = \mathbb{Z}_2 \oplus\mathbb{Z}_2$$, $$H = \mathbb{Z}_2 \oplus 0 = \langle (1,0)\rangle$$ and $$\tau: \mathbb{Z}_2 \oplus \mathbb{Z}_2 \to \mathbb{Z}_2 \oplus \mathbb{Z}_2: (x,y) \mapsto (y,x+y).$$ Then $$\tau(H) = 0 \oplus \mathbb{Z}_2 = \langle (0,1)\rangle$$ and $$\tau^2(H) = \langle (1,1)\rangle \neq H.$$