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.

  • $\begingroup$ Did you check the symmetric group $S_6$ with an outer automorphism? [$H$ being an $S_5$; I'm not sure if the product of a $2$- and a $4$-cycle is contained in the union] $\endgroup$
    – j.p.
    Apr 15, 2013 at 13:36
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    $\begingroup$ Good idea, but it doesn't work (at least for that choice of $H$). The outer automorphism takes the conjugacy class $(12)(3456)$ to itself. $\endgroup$ Apr 15, 2013 at 13:37
  • $\begingroup$ You're right, here's a reference: en.wikipedia.org/w/… $\endgroup$
    – j.p.
    Apr 15, 2013 at 13:48
  • $\begingroup$ Not sure that this is helpful or not, but note that $x=\alpha(h)^g$ if and only if $\alpha(x)=h^{\alpha(g)}$. So the question can be rephrased as saying: does there exist an $H\lt G$, such that for every $x\in G$, either $x$ or $\alpha(x)$ is contained in one of the conjugates $H^g$? I don't know how I feel about the validity of this question, but my intuition is that a good starting place is a group where there are many small conjugacy classes, and $H$ "eats up" many of them with its conjugates. $\endgroup$
    – user641
    Apr 15, 2013 at 21:36
  • $\begingroup$ Noting that $\alpha$ must clearly be an outer automorphism, we see that it is equivalent to ask if $G=\bigcup_{\theta\in \mathcal{I}\cup\mathcal{I}\alpha}H^\theta=\bigcup_{\theta\in A}H^\theta$ where $\mathcal{I}=\operatorname{Inn}(G)$ and $A$ is the complete inverse image of $\langle \alpha \mathcal{I}\rangle$ in $\operatorname{Aut}(G)$. Since you noted that $\bigcup_{\theta\in\operatorname{Aut}(G)}H^\theta$ can sometimes be a group, it makes sense to look for examples in the finite simple groups, as many of them satisfy $|\operatorname{Out}(G)|=2$. $\endgroup$
    – Alexander Gruber
    Apr 15, 2013 at 23:07

1 Answer 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.

  • $\begingroup$ Hi Peter, welcome to math.stackexchange! Why does a positive answer require the classification? Shouldn't an example be enough? (Unfortunately I can't access the link to verify if you meant "The answer is no".) $\endgroup$
    – j.p.
    Jun 14, 2013 at 12:16
  • $\begingroup$ Thanks, sorry, as you apparently expected I meant to say `no'. Fixed now. $\endgroup$ Jun 14, 2013 at 12:26
  • $\begingroup$ Thanks! I added a CW answer back at the earlier question pointing to this one. $\endgroup$ Jun 14, 2013 at 14:33

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