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I am being asked to show an example of when this fails or prove it rigorously. I am thinking of using an example to disprove the claim that a group can be a union of three subgroups. However I am not so sure if proving it rigorously might be a better way. What do you guys think?

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    $\begingroup$ Give an example of a small group that is the union of three proper subgroups. $\endgroup$ Oct 5, 2014 at 22:05
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    $\begingroup$ Okay so there is no group that exists then. Would the Klein V group work? $\endgroup$
    – cambelot
    Oct 5, 2014 at 22:06
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    $\begingroup$ The Klein V group can be expressed as the union of three of its subgroups. It shouldn't be too hard for you to find this expression, especially since there are only three non-trivial proper subgroups of it. $\endgroup$ Oct 5, 2014 at 22:08
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    $\begingroup$ @graydad: I think a better way to make the question interesting would be (as Daniel Fischer implicitly suggested) that the subgroups are to be proper rather than distinct. Otherwise if there is any non-trival subgroup $H$ we would have $\{e\}\cup H\cup G=G$ with distinct summands. $\endgroup$ Oct 5, 2014 at 22:25
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    $\begingroup$ the quaternion group of order $8$ is $<i> \cup <j> \cup <k>$ $\endgroup$ Oct 5, 2014 at 22:59

2 Answers 2

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The Klein 4-group $V_4$ is the union of three proper subgroups. A group cannot be the union of two of its proper subgroups. It might be interesting to know, that research has been done on how this might generalize.

Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $V_4$.

The proof appeared in the American Math. Monthly $77$, no. $1 (1970)$. The theorem seems to be proved earlier by the Italian mathematician Gaetano Scorza, I gruppi che possono pensarsi come somma di tre loro sottogruppi, Boll. Un. Mat. Ital. $5 (1926), 216-218$.

For 4, 5 or 6 subgroups a similar theorem is true and the Klein 4-group is for each of the cases replaced by some finite set of groups. For 7 subgroups however, it is not true: no group can be written as a union of 7 of its proper subgroups. This was proved by Tomkinson in 1997.

There is a nice overview paper by Mira Bhargava, Groups as unions of subgroups, The American Mathematical Monthly, $116$, no. $5, (2009)$.

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this may not be much help, but there is a numeric constraint which might serve as a filter (if my understanding of the question is correct)

if the group is of finite order $g$, the subgroups $g_1,g_2,g_3$, their intersections $g_{12},g_{23},g_{31},g_{123}$ then we require:

$$ g-g_1-g_2-g_3+g_{12}+g_{23}+g_{31} - g_{123} = 0 $$ subject to the constraints of the division lattice implied by Lagrange's theorem.

the quaternion group gives: $$ 8 -4-4-4+2+2+2-2 =0 $$

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