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Let $H_{1}, H_{2}$ be subgroups of $G$ satisfying, $H_{1}\leq H_{2}\leq G$. It is easy to prove that $\bigcup_{i=1}^{2}H_{i}$ is a subgroup of G. Consider an infinite ascending chain of subgroups of G, i.e. $H_{1}\leq H_{2}\leq \cdots$.

I know that by fixing an $n\in \mathbb{N}$ we can show that the union of the first $n$ subgroups is again a subgroup of $G$. Is it meaningful to talk about the union of such groups? And is it possible to extend this to the infinite case?

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It is definitely meaningful. Taking an infinite union is no problem, and the result is a subgroup. –  Grumpy Parsnip Oct 13 '11 at 2:08
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You don't even need the subgroups to form a chain. It is enough that your family $\{H_i\}_{i\in I}$ satisfy that for all $i,j\in I$ there exists $k\in I$ such that $H_i,H_j\subseteq H_k$ (that is, it is a directed family). Then $\cup H_i$ is a subgroup. –  Arturo Magidin Oct 13 '11 at 3:37
    
And more generally, if you have an algebra (in the sense of universal algebra) of type $\Omega$, and if the supremum of the arities of the operations in $\Omega$ is $\kappa$, and if $\{B_i\}_{i\in I}$ is a family of subalgebras with the property that for every subfamily $\{B_j\}_{j\in J}$ with $J\subseteq I$ and $|J|\leq\kappa$ there exists $i_J\in I$ such that $B_j$ is a subalgebra of $B_{i_J}$ for each $j\in J$, then $\cup B_i$ is a subalgebra. –  Arturo Magidin Oct 13 '11 at 3:37

4 Answers 4

up vote 6 down vote accepted

More generally, let $I$ be an index set of arbitrary non-zero cardinality, and suppose that for every $\alpha\in I$, $H_\alpha$ is a subgroup of $G$.

Suppose also that for any $\alpha, \beta \in I$, we have $H_\alpha \subseteq H_\beta\,$ or $\,H_\beta \subseteq H_\alpha$. Then $\bigcup_\alpha H_\alpha$ is a subgroup of $G$.

The question that was asked has $I=\mathbb{N}$, and $H_m \subseteq H_n$ if $m \le n$.

The proof is straightforward. We first show that if $u$ and $v$ are in $\bigcup_\alpha H_\alpha$, then $uv\in \bigcup_\alpha H_\alpha$.

Since $u \in \bigcup_\alpha H_\alpha$, there is an index $\beta$ such that $u \in H_\beta$. Similarly, $v \in H_\gamma$ for some $\gamma\in I$. By our condition on the $H_\alpha$, we have $H_\beta \subseteq H_\gamma$, or $H_\gamma \subseteq H_\beta$. Without loss of generality we may assume that $H_\beta \subseteq H_\gamma$. Then $u$ and $v$ are both in $H_\gamma$, and therefore so is their product $uv$. It follows that $uv \in \bigcup_\alpha H_\alpha$.

The other things we need to check are easier. The unit element (of $G$) is in all the $H_\alpha$, so it is in their union. Also, if $x \in \bigcup_\alpha H_\alpha$, then $x \in H_\beta$ for some $\beta$, so $x^{-1} \in H_\beta$, and therefore $x^{-1} \in \bigcup_\alpha H_\alpha$.

Comment: The condition that for all $\alpha, \beta \in I$, we have $H_\alpha \subseteq H_\beta\,$ or $\,H_\beta \subseteq H_\alpha$ can be weakened. Everything goes through if for all $\alpha, \beta \in I$, there is a $\gamma \in I$ such that $H_\alpha \subseteq H_\gamma\,$ and $\,H_\beta \subseteq H_\gamma$.

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Yes. The union of an infinite chain of subgroups is again a subgroup. This also works for subrings, subspaces, in fact, most every sub-thing.

Suppose $H_i$ is a subgroup of $G$ for each $i=1,2,\dots$ and in addition suppose that $H_i \subseteq H_{i+1}$ for each $i$.

Let $H=\cup_{i=1}^\infty H_i$. This is certainly a nonempty subset of $G$. Next, let $a,b \in H$. This means $a \in H_i$ for some $i$ and $b \in H_j$ for some $j$. Either $i<j$ and so $H_i \subseteq H_j$ and thus both $a$ and $b$ are in $H_j$ or vice-versa: $j<i$ and both elements are in $H_i$. Suppose both are in $j$. Then so is $ab$ and $a^{-1}$ since $H_j$ is a subgroup. Therefore, $ab,a^{-1} \in H_j \subseteq \cup_{k=1}^\infty H_k=H$. Therefore, $H$ is a subgroup of $G$.

It's easy to see why this type of argument should work in many different contexts. Also, note that the index set "$1,2,\dots$" can be replaced with any totally ordered set or partially ordered sets such that for all $i,j$ there is $k$ such that $i,j \leq k$.

Here's an example: We can stick $S_n$ inside $S_m$ for any $n<m$. All of these symmetric groups can be viewed as subgroups of $S_\infty$ (permutations of $\mathbb{Z}_{>0}={n \in \mathbb{Z} \,|\, n \geq 0 }$). The subgroup we get $S = \cup_{i=1}^\infty S_i$ is the set of all permutations fixing all but finitely many positive integers.

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This is an example of a rather general phenomenon. Let $G$ be any group, and let $\mathscr{H}$ be a chain of subgroups of $G$, meaning that for any $H_0,H_1\in\mathscr{H}$, either $H_0\subseteq H_1$ or $H_1\subseteq H_0$. Then $H=\bigcup\mathscr{H}$ is a subgroup of $G$.

Certainly $H\subseteq G$. Now suppose that $h_0,h_1\in H$; to show that $H \le G$, we must show that $h_0h_1,h_0^{-1}\in H$. Since $h_0\in H$, there must be some $H_0\in\mathscr{H}$ such that $h_0\in H_0$. $H_0$ is a subgroup of $G$, so $h_0^{-1}\in H_0 \subseteq H$, as desired. Similarly, there must be some $H_1\in\mathscr{H}$ such that $h_1\in H_1$. Since $\mathscr{H}$ is a chain, either $H_0 \subseteq H_1$ or $H_1 \subseteq H_0$. Without loss of generality $H_0 \subseteq H_1$, in which case $h_0,h_1\in H_1$. But $H_1$ is a subgroup of $G$, so $h_0h_1\in H_1 \subseteq H$, and the proof that $H\le G$ is complete.

Notice that what made this work is that each of the conditions that had to be checked involved only finitely many elements of $H$. Specifically, they were closure conditions of the following form:

$$\text{If }h_0,\dots,h_n \in H,\text{ then }\phi(h_0,\dots,h_n)\in H.\tag{1}$$

Suppose that $S$ is any structure with the property that a subset $H$ of $S$ is a substructure iff it satisfies a list of condition of the form $(1)$. Then if $\mathscr{H}$ is a chain of substructures of $S$, and $H = \bigcup\mathscr{H}$, the same kind of argument that I used for the group $G$ will show that $H$ is a substructure of $S$:

If $h_0,\dots,h_n\in H$, there are substructures $H_0,\dots,H_n \in \mathscr{H}$ such that $h_k\in H_k$ for $k=0,\dots,n$. The $H_k$ with $k=0,\dots,n$ are linearly ordered by $\subseteq$, so one of them contains all the rest. Let that one be $H_i$. Then $h_0,\dots,h_n\in H_i$, and since $H_i$ is a substructure of $S$, it’s closed under the function $\phi$, and hence $\phi(h_0,\dots,h_n)\in H_i \subseteq H$.

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This is possible, and the way to do it is made precise using the notion of a colimit (more specifically, a direct limit). This is a general categorical construction, and gives you a way to take a diagram of groups, and spit out another group in a meaningful way.

In your case, the diagram we are looking at is $H_1\to H_2\to H_3\to\cdots$, where the arrows are the inclusion maps. The colimit of this diagram will be a group $G$ such that we have maps $H_i\to G$ for all $i$ such that the big diagram containing our original diagram of $H_i$'s together with all the maps into $G$ commutes. It is universal in the sense that if we have any other set of maps $H_i\to K$ such that the same big diagram commutes, then there is a unique map $G\to K$ making everything commute.

In the category of sets, if we have inclusions $X_1\subseteq X_2\subseteq X_3\subseteq\cdots$, then the colimit of this diagram is $\bigcup_{i=1}^\infty X_i$. It is worth noting that if each of your groups $H_i$ is a subgroup of some group $G$, then as a set, this colimit will just be the union.

If you want an example of such a colimit where all the groups don't live inside some larger group, consider the inclusions $$\mathrm{GL}_1(F)\to\mathrm{GL}_2(F)\to\mathrm{GL}_3(F)\to\cdots$$ where the arrows are inclusions are into the upper left hand corner. The colimit of this diagram is $\mathrm{GL}_\infty(F)$, and appears in algebraic topology.

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@SL_{2}, how can I think about this in elementary abstract algebra terms? –  Edison Oct 13 '11 at 2:21
    
@ElG As I said (and as other answers have elaborated upon), if each of the $H_i$'s are subgroups of some bigger group $G$, then the colimit will just be the union. If you're not familiar with category theory, then this is probably the best way to think about it. –  SL2 Oct 13 '11 at 2:29

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