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"Let $G$ be a group, and suppose $G=H \cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$."

Let $h \in H$ and $k \in K$. Then $hk \in H$ or $hk \in K$ (since every element of $G$ is in either $H$ or $K$). If $hk=h'$ for some $h' \in H$, then $k=h^{-1}h'$, so $k \in H$. If $hk=k'$ for some $k' \in K$, then $h=k'k^{-1}$ so that $h \in K$.

If for all $h \in H$ we have $h \in K$, or if for all $k \in K$ we have $k \in H$, then $H \subseteq K$ or $K \subseteq H$. Then since $G=H \cup K$, we must have either $H=G$ or $K=G$.

I'm not sure if the first paragraph of my 'proof' implies the second. I've shown that for arbitrary $h \in H$, $h \in H$ and possibly $h \in K$, and similar for $k \in K$. I don't know how to wrap it up (or perhaps this route won't lead anywhere at all).

If this way won't work, I'd just like a hint on a new direction to take.

Thanks.

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    $\begingroup$ The problem with trying to go from the first paragraph to the second is that you know that $hk\in H$ or $hk\in K$ for every $h\in H$ and $k\in K$, but you cannot distribute this "for every" across the logical disjunctive (the "or"); that is, $$\forall h\in H,k\in K: (hk\in H\vee hk\in K)$$ does not logically entail $$(\forall h\in H,k\in K:hk\in H)\vee(\forall h\in H,k\in K:hk\in K),$$ and it is the latter that creates the premises for your second paragraph. $\endgroup$
    – anon
    Mar 19, 2013 at 3:08
  • $\begingroup$ It follows easily from the answers given in this topic. $\endgroup$
    – user26857
    Apr 6, 2013 at 9:20
  • $\begingroup$ @anon , Could you give an example when the first statement is true but the second is false? it's tricky for me, how the first doesn't imply the other ! $\endgroup$
    – Fawzy Hegab
    Sep 4, 2013 at 23:30
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    $\begingroup$ @FawzyHegab It seems my $1$st comment makes too strong of a claim. My implicit broader point - that "for all" does not in general distribute over "or" and thus needs to be justified specially in this instance - remains true. Here's a proof of this particular implication: the first statement says that $HK\subseteq H\cup K$. But $H\cup K\subseteq HK$, so $HK=H\cup K$ (suppose true). The second statement is equivalent to "$H\subseteq K$ or $K\subseteq H$"; suppose this is false. If $h\in H-K,k\in K-H$ then wlog $hk=h'\in H$ implies $k=h^{-1}h'\in (K\cap H)\cap(K-H)=\varnothing$, absurd. $\endgroup$
    – anon
    Sep 6, 2013 at 3:49
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    $\begingroup$ @FawzyHegab Assume for the sake of argument that all people are either male or female (in reality, humans and organisms in general are too diverse and varied for his to be an uncomplicated universal truth). Then "for all people x, x is male or x is female" is true, but "all people are male or all people are female" is false. $\endgroup$
    – anon
    Sep 6, 2013 at 21:42

2 Answers 2

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Suppose both $H,K$ are distinct and proper in $G$. Then pick $h\in H\setminus K$ and $k\in K\setminus H$.

In which of $K$ or $H$ or both does $hk$ lie?

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    $\begingroup$ Doesn't $hk$ lie in neither $H$ nor $K$? If it was in one of them, the argument in my second paragraph above would create a contradiction with our choice of $h \in H \setminus K$ and $k \in K \setminus H$. Then $hk \not\in H$ and $hk \not\in K$, so $hk \not\in H \cup K$, so that $G \ne H \cup K$, and we have proved the contrapositive. $\endgroup$
    – Alex Petzke
    Mar 19, 2013 at 21:11
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    $\begingroup$ @AlexPetzke Exactly. $\endgroup$
    – anon
    Mar 19, 2013 at 21:15
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Bonus information: as you noted, 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 $C_2 \times C_2$.

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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    $\begingroup$ This is very interesting! (+1) $\endgroup$
    – Shaun
    Jul 10, 2022 at 13:58

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