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The union of an ascending chain of subgroups is a subgroup

ianchenmu

Member
Feb 3, 2013
74
Let $G$ be a group, and $\left \{ H_{i} \right \}_{i\in \mathbb{Z}}$ be an ascending chain of subgroups of $G$; that is, $H_{i}\subseteq H_{j}$ for $i\leqslant j$. Prove that $\bigcup _{i\in \mathbb{Z}}H_{i}$ is a subgroup of $G$.

I don't need the proof now. But can you show an example for me that the assertion fails to a set of subgroups that do not satisfy the ascending chain condition?
 
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jakncoke

Active member
Jan 11, 2013
68
Let $G$ be a group, and $\left \{ H_{i} \right \}_{i\in \mathbb{Z}}$ be an ascending chain of subgroups of $G$; that is, $H_{i}\subseteq H_{j}$ for $i\leqslant j$. Prove that $\bigcup _{i\in \mathbb{Z}}H_{i}$ is a subgroup of $G$.






I don't need the proof now. But can you show an example for me that the assertion fails to a set of subgroups that do not satisfy the ascending chain condition?
Take, $a,b \in \bigcup_{i \in \mathbb{Z}}H_i$, then let $p \in \mathbb{Z}$ be the index of a subgroup containting a, and $j \in \mathbb{Z}$ be the index of a subgroup containing b(Clearly one should exist). Now since Z is well ordered. Take $max(p, j) = k$ or if p = j, max(p,j) = p =j=k. So clearly, since this is an ascending chain of subgroups the subgroup $H_k$ contains both a,b. Since $H_{k}$ is a subgroup it contains $b^{-1}$, and is closed under group operation so $ab^{-1} \in H_{k}$ and since $H_k \subset \bigcup_{i \in \mathbb{Z}} H_{i}$ and thus also $ab^{-1} \in \bigcup_{i \in \mathbb{Z}} H_{i}$. Thus $\bigcup_{i \in \mathbb{Z}} H_{i}$ is a subgroup of G
 
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ianchenmu

Member
Feb 3, 2013
74
Take, $a,b \in \bigcup _{i\in \mathbb{Z}}H_{i}$ , then let $i \in \mathbb{Z}$ be the index of the subgroup containting a, and $j \in \mathbb{Z}$ be the index of the subgroup containing b. Now since Z is well ordered. Take $max(i, j) = k$ or if i = j, max(i,j) = i =j=k. So clearly, since this is an ascending chain of subgroups the subgroup $H_j$ contains both a,b. Since H_j is a subgroup it contains $b^{-1}$, and is closed under group operation so $ab^{-1} \in H_{k}$ and thus also $\in \bigcup _{i\in \mathbb{Z}}H_{i}$ since $H_{k} \subset \bigcup _{i\in \mathbb{Z}}H_{i}$
Thank you but there seems something wrong with your use of LaTex in the reply so I can't read it.
 

jakncoke

Active member
Jan 11, 2013
68
now you should be able to.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
It is easy to find counter-examples when the ascending chain condition is not satisfied.

For example, let $G = \Bbb Z$ and define:

$H_1 = 2\Bbb Z$
$H_2 = H_{2+i} = 3\Bbb Z$ for all $i \in \Bbb N$.

(if one wants to index over the integers, set $H_j = \{0\}$ for all non-positive j).

Then $\bigcup_{i \in \Bbb N} H_i = 2\Bbb Z \cup 3\Bbb Z$, which is not closed under addition, for example, it does not contain 5 = 2+3.

In fact, analogous to the situation with vector subspaces, in general for two subgroups, $H,K \subseteq G$, we have that $H \cup K$ is NOT a subgroup, unless either:

$H \subseteq K$ or $K \subseteq H$.

To see this, suppose we have two subgroups $H,K$ of $G$, and that $K$ is not a subset of $H$, but $H \cup K$ is a subgroup.

Then there exists $k \in K - H$. Now, let $h$ be any element of $H$. Since $h,k \in H \cup K$, which is a subgroup of $G$, by closure we have $hk \in H \cup K$. Thus either:

$hk \in H$ or $hk \in K$.

If $hk \in H$, then since $H$ is a subgroup $h^{-1} \in H$, and thus:

$k = h^{-1}(hk) \in H$, contradicting our choice of $k$.

Thus we MUST have $hk \in K$, and since $K$ is a subgroup, $k^{-1} \in K$, so that:

$h = (hk)k^{-1} \in K$. Since $h$ was chosen arbitrarily, this means $H \subseteq K$.

For reasons that are not quite clear to me, "union" behaves poorly with regards to most algebraic operations, whereas "intersection" usually behaves quite well. In general, for algebraic objects $A,B$ the union isn't "big enough" to be an object itself: a good example is the commutator subgroup $[G,G]$ of a group $G$: often the commutators $[a,b] = aba^{-1}b^{-1}$ aren't "enough" to make a subgroup, we have to include all arbitrary PRODUCTS of commutators as well, to ensure closure. This phenomenon occurs in various algebraic realms:

Groups
Abelian Groups
Rings
Fields
$R$-modules
Vector Spaces
$k$-algebras

In each case $\langle A,B \rangle$ is usually much larger than $A \cup B$ (although it contains it).

For groups, to get the subgroup $\langle H,K \rangle$, we often have to go quite a big bigger than the union, or even the product $HK$ (if $HK \neq KH$ even this is not large enough). So to get a group that contains both $H$ and $K$, we often seek to impose restrictions on "what type of subgroups they are", such as the ascending chain condition listed here (I hope you can see how the discussion of two subgroups generalizes to an indexed family).

In a sense, the direct product $H \times K$ represents "an extreme case" where $H$ and $K$ "don't interact at all". This puts at least, an upper bound on where to look for a "group containing $H$ and $K$ as subgroups". Because groups can be non-abelian, it is possible to have "intermediate stages" in-between the two conditions:

$H \subseteq K = G$
$H \times K = G$

such as a semi-direct product, for example. If neither $H$ nor $K$ are normal subgroups, finding the subgroup generated by their union can be a difficult task (somewhat easier if $G$ is finite, where one can laboriously examine each possible subgroup in turn).
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Take, $a,b \in \bigcup_{i \in \mathbb{Z}}H_i$, then let $p \in \mathbb{Z}$ be the index of a subgroup containting a, and $j \in \mathbb{Z}$ be the index of a subgroup containing b(Clearly one should exist).
Hello Jack.
I think here you are presuming that the index of the subgroup containing $a$ is an integer which I don't think is necessarily true. For example if we consider $\mathbb Q$ as a subgroup of $\mathbb R$ then we have that $[\mathbb R:\mathbb Q]$ is not finite.
 

jakncoke

Active member
Jan 11, 2013
68
Hello Jack.
I think here you are presuming that the index of the subgroup containing $a$ is an integer which I don't think is necessarily true. For example if we consider $\mathbb Q$ as a subgroup of $\mathbb R$ then we have that $[\mathbb R:\mathbb Q]$ is not finite.
I don't think i mean "index" in the same way you are thinking, you are thinking of the "relative size" of Q in R, which confusingly enough is also called an index. My index refers to a subgroup (like a label or a pointer) in the chain. Where did i get this notion? It was stated by the author that $H_{i \in \mathbb{Z}}$ is an ascending chain.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
I don't think i mean "index" in the same way you are thinking, you are thinking of the "relative size" of Q in R, which confusingly enough is also called an index. My index refers to a subgroup (like a label or a pointer) in the chain. Where did i get this notion? It was stated by the author that $H_{i \in \mathbb{Z}}$ is an ascending chain.
Oh! No problem then.