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Let $G$ be a group and $\{H_i\}_{i\in I}$ a family of subgroups. I would like to find a condition that will imply that $\cup_{i\in I} H_i$ is a subgroup. I know that it's not true in general, I need help to find this condition.

Any help is welcome.

Thanks a lot.

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    $\begingroup$ are you interested in an equivalent condition for the union being a group, or just any condition at all that will just guarantee the union is a group? (for the latter, how about the condition that all $H_i$ are equal?) $\endgroup$ – Ittay Weiss Apr 21 '13 at 7:47
  • $\begingroup$ @IttayWeiss just a good condition to imply that the union is a subgroup is enough, nonetheless a good equivalent condition would be welcome also :) $\endgroup$ – user42912 Apr 21 '13 at 7:51
  • $\begingroup$ so can you give any criterion for good conditions? Is the condition I just supplied in my comment good? $\endgroup$ – Ittay Weiss Apr 21 '13 at 7:57
  • $\begingroup$ @IttayWeiss I think the condition of Suugaku is better because your condition is so restricted. $\endgroup$ – user42912 Apr 21 '13 at 8:05
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This result is well know for two subgroup $$H_1\cup H_2\,\,\text{is a subgroup}\iff H_1\subset H_2\,\,\text{or}\,\, H_2\subset H_1$$ indeed suppose that $H_1\not\subset H_2$ and $ H_2\not\subset H_1$ then there's $h_1\in H_1\setminus H_2$ and $h_2\in H_2\setminus H_1$ and let $z=h_1+h_2$ so $z\in H_1\cup H_2$ which is a goup so if $z\in H_1$ then $h_2=z-h_1\in H_1$ is a contradiction and the same conclusion if $z\in H_2$.

Edit Note we can easily generalize this result proving this condition: there's $i_0\in I$ such that $H_i\subset H_{i_0}$ for every $i\in I$ is sufficient to have $\cup_I H_i$ is a subgroup.

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    $\begingroup$ What kind of generalization did you have in mind? $\endgroup$ – Mikko Korhonen Apr 21 '13 at 7:33
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    $\begingroup$ What about Klein group? $\endgroup$ – Boris Novikov Apr 21 '13 at 7:38
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    $\begingroup$ The problem with your generalization is that for example, when $H_1 \cup H_2 \cup H_3$ is a subgroup, it does not necessarily follow that $H_2 \cup H_3$ is a subgroup. $\endgroup$ – Mikko Korhonen Apr 21 '13 at 7:57
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    $\begingroup$ @IttayWeiss Can I take a few minutes to reflection. $\endgroup$ – user63181 Apr 21 '13 at 8:03
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    $\begingroup$ @user42912 You're welcome. $\endgroup$ – user63181 Apr 21 '13 at 8:44
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Generally speaking is is very rare that a union of finitely many subgroups is again a subgroup. As a consequence the various sufficient conditions that one can formulate for a union of subgroups to be a subgroup generally amount to boring truisms for the case of finitely many subgroups. Examples are "there is one $H_i$ that contains all the others" (the union will be equal to $H_i$), or "every element of $G$ lies in at least one subgroup $H_i$" (the union will be all of $G$). The example of taking for $(H_i)_{i\in I}$ the collection of all cyclic subgroups of $G$ (which will work, with the union being $G$, whether or not this collection is finite) shows that it will be hard to formulate any nice necessary condition for $\bigcup_iH_i$ to be a subgroup (note that leaving out just one of those cyclic subgroups will often destroy the property that $\bigcup_iH_i$ is a subgroup).

If your collection of subgroups is infinite, then there are some interesting sufficient conditions that can be formulated. Notably "for every pair of indices $i_1,i_2\in I$ there exists $j\in I$ such that $H_{i_1}\subseteq H_j$ and $H_{i_2}\subseteq H_j$" (in other words, ordering the collection of subgroups by inclusion, every pair has an upper bound) suffices for $\bigcup_iH_i$ to be a subgroup. Indeed, if $x,y\in\bigcup_iH_i$ then there exist $i_1,i_2\in I$ with $x\in H_{i_1}$ and $y\in H_{i_2}$, and with $j$ as in the requirement one has $xy\in H_j\subseteq\bigcup_iH_i$ (that the union contains the identity and is closed under inverses needs no condition in the $H_i$ at all, except $I\neq\emptyset$ which certainly should be assumed). Note that the condition is a valid sufficient condition also for finite families of subgroups, but it is a boring one, as in that case it can easily be shown to imply that among the subgroups there is one that contains them all.

Note also that a special instance of this condition is when the set of subgroups is totally ordered by inclusion (for every pair, one of them contains the other), in which case $j$ can be chosen among $\{i_1,i_2\}$; this gives the sufficient condition ("$H_j<H_j$ or $H_j<H_i$") mentioned in other answers.

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    $\begingroup$ The sufficent condition you mention says that the poset of subgroups (ordered by inclusion) is a directed set. en.wikipedia.org/wiki/Directed_set $\endgroup$ – Mikko Korhonen Apr 21 '13 at 9:55
  • $\begingroup$ @m.k. Right, thank you. I was not acquainted with that term. $\endgroup$ – Marc van Leeuwen Apr 21 '13 at 10:01
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The condition to impose is that $H_i < H_j$ or $H_j < H_i$ for all $i,j \in I$. If this is true, then one can check that the union of all $H_i$ is a subgroup. If $x,y \in \cup H_i$, then $x \in H_i$ for some $i$ and $y \in H_j$ for some $j$. Since $H_i < H_j$ or $H_j < H_i$, it must be the case that $x,y \in H_k$ for one of $k = i$ or $k = j$. Since $H_k$ is a subgroup, it means that $xy^{-1} \in H_k \subset \cup H_i$ proving that $\cup H_i$ is a subgroup.

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  • $\begingroup$ how do you propose to prove the other implication? (hint, you can't: in any non-cyclic group $G$, take the family of all subgroups generated by a single element. The union is a subgroup, but the condition you present fails. $\endgroup$ – Ittay Weiss Apr 21 '13 at 7:39
  • $\begingroup$ As far as I can tell, the OP only wants this implication. I never said anything about it being equivalent. However, thanks for giving the example to prove that the other implication is not true. It is interesting. If we want the condition to have be an "if and only if" I'm not sure if there is such a one? $\endgroup$ – Suugaku Apr 21 '13 at 7:42
  • $\begingroup$ I see. Fair enough. $\endgroup$ – Ittay Weiss Apr 21 '13 at 7:46
  • $\begingroup$ @IttayWeiss: not that it really matters, but I think there are non-cyclic groups where all subgroups generated by a single element are nested. I'm thinking of the group of dyadic (i.e. denominator a power of 2) rationals quotiented by the integers. $\endgroup$ – Ben Millwood Apr 21 '13 at 7:51
  • $\begingroup$ @BenMillwood nope. Take any non-trivial element, let $H_1$ be generated by it. Now, take any element not in $H_1$, let $H_2$ be generated by it. Continue (transfinitely if needed) this way to obtain obviously non-nested subgroups whose union is the entire group. $\endgroup$ – Ittay Weiss Apr 21 '13 at 7:56

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