Covering a group with the conjugates of two subgroups related by an automorphism

Let $G$ be a finite group and $H$ a proper subgroup. Then $G$ is not the union of the conjugates of $H$. This is a standard homework problem; Arturo gives a nice solution here.

It is also not possible for $G$ to be $H_1 \cup H_2$ with $H_1$ and $H_2$ both proper subgroups, by a simple counting argument.

It is possible to have $G = \bigcup_g g H_1 g^{-1} \cup \bigcup_g g H_2 g^{-1}$. Consider, for example $G=S_3$, $H_1 = A_3$ and $H_2 = S_2$. See this paper for results on covering $S_n$ and $A_n$ in this way, and this paper for covering $GL(n, q)$.

It is also possible that $G = \bigcup_{\alpha \in \mathrm{Aut}(G)} \alpha(H)$. For example, $G = (\mathbb{Z}/p)^2$ and $H = \mathbb{Z}/p$.

Is it possible that $G$ is a finite group, $\alpha$ an order $2$ automorphism, and $H$ a proper subgroup, such that $G= \bigcup_{g \in G} g H g^{-1} \cup \bigcup_{g \in G} g \alpha(H) g^{-1}$?

This is essentially the group theory question that arises if you try to answer this question. More precisely, that question doesn't need $\alpha^2$ to be the identity, it just needs $\alpha^2$ to be inner.

Solution 1:

The answer is no, by a result of Jan Saxl which requires the classification of the finite simple groups, and a previous result of Klingen which reduces matters to a question about simple groups. See http://dx.doi.org/doi:10.1112/jlms/s2-38.2.243 for Saxl's paper.