MHB Proving Finite Groups: A Contradiction in Subgroups - Hello! (Wave)

[evinda]

Hello! (Wave)

I want to show that a group with finite number of subgroups has to be finite.

I thought to suppose that the group is not finite and then get a contradiction.So, suppose that the group is not finite.

Then the group has infinite elements.

Since the group has a finite number of subgroups, the infinite elements of the group have to be shared to a finite number of subgroups. Contradiction.Is this right? Or could we justify it better?

 

[Deveno]

The first thing you want to show is that every element of $G$ must have finite order.

(Why? Well, $\Bbb Z$ has an infinite number of subgroups).

The next thing you need to show is that:

$G = \bigcup\limits_{g \in G} \langle g\rangle$.

By assumption, the union on the right must be a finite one of finite groups.

(EDIT: By the way, it is *not* true that an infinite group must have elements of infinite order, one can take the group of (countably) infinite sequences consisting of $0$'s and $1$'s, adding term-by-term modulo $2$, in which EVERY element has order two).

(EDIT #2: The union of a finite number of subsets need not be finite-if anyone of the subsets is infinite).

 

[evinda]

The first thing you want to show is that every element of $G$ must have finite order.

So in order $G$ to be finite does it have to have finite order?
Also in order $G$ to have finite order , does every element of $G$, so every subgroup of $G$ have to have finite order?

(Why? Well, $\Bbb Z$ has an infinite number of subgroups).

Does this imply that $\Bbb Z$ is infinite, so it has infinite order?

The next thing you need to show is that:

$G = \bigcup\limits_{g \in G} \langle g\rangle$.

$G$ can be written in this way if we know that the subgroups are cyclic, or not? (Thinking)
Does this hold in this case?

 

[Deveno]

So in order $G$ to be finite does it have to have finite order?
Also in order $G$ to have finite order , does every element of $G$, so every subgroup of $G$ have to have finite order?



Does this imply that $\Bbb Z$ is infinite, so it has infinite order?



$G$ can be written in this way if we know that the subgroups are cyclic, or not? (Thinking)
Does this hold in this case?

A group of finite order, by definition, is a finite set, since the order of a group is the cardinality of its underlying set.

$\Bbb Z$ is a cyclic group of infinite order (that is, it is generated by a single element, and it has infinitely many elements). It also has infinitely many subgroups, for every $k \in \Bbb Z^{+}$, we have the subgroup $k\Bbb Z$, these are all distinct.

Now any ELEMENT $a$ of infinite order generates an infinite subgroup of $G$ (namely $\langle a\rangle$) that is isomorphic to $\Bbb Z$, and thus $G$ would have an infinite number of subgroups: $\langle a\rangle, \langle a^2\rangle,\langle a^3\rangle,\dots$ etc.

Since by assumption "our" $G$ has only a FINITE number of subgroups, it cannot have any elements of infinite order, thus all its elements are of finite order, that is: $|\langle g\rangle| < \infty$ for any $g \in G$ (the order of an element is the order of the cyclic subgroup it generates).

It is true that for ANY group $G$, that:

$G = \bigcup\limits_{g \in G} \langle g\rangle$

since for any SET, $X$, we have:

$X = \bigcup\limits_{x \in X} \{x\}$

and certainly, we have $\bigcup\limits_{g \in G} \{g\} \subseteq \bigcup\limits_{g \in G} \langle g\rangle \subseteq G$, since:

$g \in \langle g\rangle$, for each $g \in G$.

I'll repeat this, for emphasis:

EVERY group is the union of its cyclic subgroups.

So if a group has only a finite number of subgroups, it must have only a finite number of cyclic subgroups, and if each of these cyclic subgroups is finite (which they must be, since an infinite cyclic group has an infinite number of subgroups), then $G$ is a finite union of finite sets (the cyclic subgroups), which is therefore, finite.

 

[evinda]

$\Bbb Z$ is a cyclic group of infinite order (that is, it is generated by a single element, and it has infinitely many elements). It also has infinitely many subgroups, for every $k \in \Bbb Z^{+}$, we have the subgroup $k\Bbb Z$, these are all distinct.

Does it hold that the infinitely many elements of $\mathbb{Z}$ are equal to all the infinitely many subgroups of $\mathbb{Z}$ ?

Now any ELEMENT $a$ of infinite order generates an infinite subgroup of $G$ (namely $\langle a\rangle$)

Could you give me an example of a group where we can see that the above statement holds?

Now any ELEMENT $a$ of infinite order generates an infinite subgroup of $G$ (namely $\langle a\rangle$) that is isomorphic to $\Bbb Z$, and thus $G$ would have an infinite number of subgroups: $\langle a\rangle, \langle a^2\rangle,\langle a^3\rangle,\dots$ etc.

Why is any infinite subgroup of $G$ isomorphic to $\Bbb Z$ ?
Also always if we have a generator $a$ of a group, all the subgroups are of the form $\langle a^n \rangle, n \in \mathbb{N}$, right?

the order of an element is the order of the cyclic subgroup it generates

Why does this hold?

It is true that for ANY group $G$, that:

$G = \bigcup\limits_{g \in G} \langle g\rangle$

since for any SET, $X$, we have:

$X = \bigcup\limits_{x \in X} \{x\}$

and certainly, we have $\bigcup\limits_{g \in G} \{g\} \subseteq \bigcup\limits_{g \in G} \langle g\rangle \subseteq G$, since:

$g \in \langle g\rangle$, for each $g \in G$.

So in our case do we take as known that $G=\bigcup_{g \in G} \{ g \}$ ?
If so then how do we deduce that if $g \in G$ then $g \in \langle g \rangle$ ?

EVERY group is the union of its cyclic subgroups.

So do we known that each group has cyclic subgroups, i.e. is the set of cyclip subgroups of any group always non-empty?

So if a group has only a finite number of subgroups, it must have only a finite number of cyclic subgroups, and if each of these cyclic subgroups is finite (which they must be, since an infinite cyclic group has an infinite number of subgroups), then $G$ is a finite union of finite sets (the cyclic subgroups), which is therefore, finite.

Aren't the cyclic subgroups a subset of the cyclic subgroups?
If so could you explain me the justification why the cuclic subgroups are finite?

 

[Deveno]

Does it hold that the infinitely many elements of $\mathbb{Z}$ are equal to all the infinitely many subgroups of $\mathbb{Z}$ ?

They are not EQUAL (the integer $2$ is not the subgroup of the integers consisting of all MULTIPLES of $2$, for example) but there IS a 1-1 correspondence. In particular, there is the same "number" of subgroups of $\Bbb Z$ as there are *elements* of $\Bbb Z$, namely countably infinitely many.

Could you give me an example of a group where we can see that the above statement holds?

Sure, in the real numbers under addition, any non-zero real number $r$ gives us an example-all of these are of infinite order, as there is no FINITE positive integer $n$ such that:

$r+r+\cdots+r = 0$ ($n$ summands). For if $nr = 0$ (with $n \in N$), then since $\Bbb R$ is a field, $n = 0$, or $r = 0$. SInce $r \neq 0$, we must have $n = 0$, but 0 is not positive.

In $(\Bbb R,+)$ the subgroup generated by $r$ is:

$\langle r\rangle = \{\dots,-3r,-2r,-r,0,r,2r,3r,4r,\dots\} = \Bbb Zr = \{kr: k \in \Bbb Z\}$.

This is the isomorphism with $\Bbb Z$: $kr \leftrightarrow k$.

In general, we have the *homomorphism*:

$\Bbb Z \to \langle a\rangle$, given by $k \mapsto a^k$. In fact, this mapping is surjective, but it is injective if and only if $a$ is not of finite order (if $a$ has finite order, say $n$, then $0$ and $n$ both map to $a^n = a^0 = e$).

Why is any infinite subgroup of $G$ isomorphic to $\Bbb Z$ ?

This is not true. Only infinite CYCLIC subgroups are isomorphic to $\Bbb Z$.

Also always if we have a generator $a$ of a group, all the subgroups are of the form $\langle a^n \rangle, n \in \mathbb{N}$, right?

Yes. In the integers, $a^n$ is typically written $n\cdot a$, since $a\ast b$ is typically written $a+b$ (it's a notational thing).

Why does this hold?

Normally, the order of an element $g \in G$ is defined as the least positive integer $n$ such that $g^n = e_G$.

If we adopt this convention we see that:

$\langle g\rangle$ contains at most $\{e = g^n,g,g^2,\dots,g^{n-1}\}$ (since higher powers can be reduced to these using $g^n = e$), so the order of $\langle g\rangle$ is at most $n$.

Now suppose two of these were the same, say $g^k = g^m$, for $0 \leq k < m < n$. Then:

$e = g^k(g^k)^{-1} = g^m(g^k)^{-1} = (g^m)(g^{n-k}) = g^{m+n-k} = g^{m-k+n} = g^{m-k}g^n = g^{m-k}$.

But $0 < m - k < n - k < n$, so that $m - k$ would be a positive number less than $n$ with $g^{m-k} = e$. This violates the minimality of $n$ (which is the LEAST such positive integer).

We thus must conclude that this cannot happen, so all the powers:

$\{e,g,g^2,\dots,g^{n-1}\}$ are DISTINCT, that is $\langle g\rangle$ has EXACTLY $n$ elements, that is:

The order of $g$ (as an element) is the order of $\langle g\rangle$ (as a subgroup).

So in our case do we take as known that $G=\bigcup_{g \in G} \{ g \}$ ?

This is true of any set, including those sets which also happen to be groups (every set is the union of its elements).

If so then how do we deduce that if $g \in G$ then $g \in \langle g \rangle$ ?

An alternate definition of $\langle g\rangle$ is the smallest subgroup of $G$ containing $g$. Put another way: every subgroup generated by a set of elements contains the elements that generate it. It would be rather odd to call a subgroup "generated by $g$" if it did not even contain the generator.

So do we known that each group has cyclic subgroups, i.e. is the set of cyclic subgroups of any group always non-empty?

Yes! For ANY group $G$, the subgroup generated by any single element is, by definition, cyclic. Most groups have LOTS of cyclic subgroups.

Aren't the cyclic subgroups a subset of the cyclic subgroups?
If so could you explain me the justification why the cyclic subgroups are finite?

Did you mean "the cyclic subgroups are a subset of all the subgroups"? If so, yes.

The cyclic subgroups are finite, because the elements that generate them are of finite order (see above).

 

[evinda]

$\Bbb Z$ is a cyclic group of infinite order (that is, it is generated by a single element, and it has infinitely many elements). It also has infinitely many subgroups, for every $k \in \Bbb Z^{+}$, we have the subgroup $k\Bbb Z$, these are all distinct.

Now any ELEMENT $a$ of infinite order generates an infinite subgroup of $G$ (namely $\langle a\rangle$) that is isomorphic to $\Bbb Z$, and thus $G$ would have an infinite number of subgroups: $\langle a\rangle, \langle a^2\rangle,\langle a^3\rangle,\dots$ etc.

Since by assumption "our" $G$ has only a FINITE number of subgroups, it cannot have any elements of infinite order, thus all its elements are of finite order, that is: $|\langle g\rangle| < \infty$ for any $g \in G$ (the order of an element is the order of the cyclic subgroup it generates).
I'll repeat this, for emphasis:

EVERY group is the union of its cyclic subgroups.

So if a group has only a finite number of subgroups, it must have only a finite number of cyclic subgroups, and if each of these cyclic subgroups is finite (which they must be, since an infinite cyclic group has an infinite number of subgroups), then $G$ is a finite union of finite sets (the cyclic subgroups), which is therefore, finite.

If there is an element $a$ of infinite order, then it generates an infinite subgroup of $G$, the $\langle a\rangle$.

Why is this subgroup isomorphic to $\mathbb{Z}$ ?

Is this isomorphism the only way to see that $G$ would have an infinite number of subgroups?

 

[Deveno]

If there is an element $a$ of infinite order, then it generates an infinite subgroup of $G$, the $\langle a\rangle$.

Why is this subgroup isomorphic to $\mathbb{Z}$ ?

Is this isomorphism the only way to see that $G$ would have an infinite number of subgroups?

If $a$ has infinite order (which means that the $a^k$ elements are *distinct* for differing $k$), then the mapping:

$a^k \mapsto k$ from $\langle a\rangle \to \Bbb Z$ is injective.

It is also easy to see that it is surjective, since any integer $n$ has the pre-image $a^n$.

Finally, it is a homomorphism, since, if $\phi(a^k) = k$, then:

$\phi(a^ka^m) = \phi(a^{k+m}) = k+m = \phi(a^k) + \phi(a^m)$.

Basically, any cyclic group is isomorphic to a quotient of the integers. If the generator has finite order, the resulting cyclic group is finite. If the generator has infinite order (and thus there are infinitely many "powers" of the generator), clearly we have an infinite cyclic group. As I indicated with $\phi$ above, an infinite cyclic group is isomorphic to $\Bbb Z$.

You do not *need* to use this isomorphism with the integers, it suffices to show that any element of infinite order generates a group with infinitely many subgroups:

If $a$ has infinite order, then:

$\langle a\rangle$
$\langle a^2\rangle$
$\langle a^3\rangle$ and so on, are all different.

(As an exercise, you might try to investigate what happens if $\langle a^k\rangle =\langle a^m \rangle$ for $k \neq m$).

 

[evinda]

It is true that for ANY group $G$, that:

$G = \bigcup\limits_{g \in G} \langle g\rangle$

since for any SET, $X$, we have:

$X = \bigcup\limits_{x \in X} \{x\}$

and certainly, we have $\bigcup\limits_{g \in G} \{g\} \subseteq \bigcup\limits_{g \in G} \langle g\rangle \subseteq G$, since:

$g \in \langle g\rangle$, for each $g \in G$.

$$G= \bigcup_{g \in G} \{ g \}$$

If $g \in G$ then $g \in \langle g \rangle \Rightarrow \bigcup_{g \in G} \{ g \} \subset \bigcup_{g \in G} \langle g \rangle \Rightarrow G \subseteq \bigcup_{g \in G} \langle g \rangle $

How do we get that $\bigcup_{g \in G} \langle g \rangle \subseteq G$ ?

 

[evinda]

You do not *need* to use this isomorphism with the integers, it suffices to show that any element of infinite order generates a group with infinitely many subgroups:

If $a$ has infinite order, then:

$\langle a\rangle$
$\langle a^2\rangle$
$\langle a^3\rangle$ and so on, are all different.

(As an exercise, you might try to investigate what happens if $\langle a^k\rangle =\langle a^m \rangle$ for $k \neq m$).

We suppose that $\langle a^k \rangle= \langle a^m \rangle$ for $|m| \neq |k|$.
If $a^k \in \langle a^m \rangle$ then $a^k=a^{mn}$ for some $n \in \mathbb{Z}$.
Then $k=mn$. Otherwise $1=a^{mn-k}$ and if $mn-k \neq 0$ then $a$ has a finite order. Contradiction.
So $k$ is a multiple of $m$.
Similarly we get that $m$ is a multiple of $k$. That happens only if $m=\pm k$.
So for $|m| \neq |k|$ we have $\langle a^k \rangle \neq \langle a^m \rangle$.

 

[Deveno]

We suppose that $\langle a^k \rangle= \langle a^m \rangle$ for $|m| \neq |k|$.
If $a^k \in \langle a^m \rangle$ then $a^k=a^{mn}$ for some $n \in \mathbb{Z}$.
Then $k=mn$. Otherwise $1=a^{mn-k}$ and if $mn-k \neq 0$ then $a$ has a finite order. Contradiction.
So $k$ is a multiple of $m$.
Similarly we get that $m$ is a multiple of $k$. That happens only if $m=\pm k$.
So for $|m| \neq |k|$ we have $\langle a^k \rangle \neq \langle a^m \rangle$.

Very nice!

$$G= \bigcup_{g \in G} \{ g \}$$

If $g \in G$ then $g \in \langle g \rangle \Rightarrow \bigcup_{g \in G} \{ g \} \subset \bigcup_{g \in G} \langle g \rangle \Rightarrow G \subseteq \bigcup_{g \in G} \langle g \rangle $

How do we get that $\bigcup_{g \in G} \langle g \rangle \subseteq G$ ?

If $A_i \subseteq X$ for each $i$, then $\bigcup\limits_i A_i \subseteq X$.

Proof:

Suppose $x \in \bigcup\limits_i A_i$.

Then since *by definition* $\bigcup\limits_i A_i = \{y \in X: \exists i\text{ with } y\in A_i\}$ (if $y$ is in a union of sets, it must lie in ONE of the sets (at least) we are taking the union of), we have $x \in A_i$ for some $i$.

Since for each (and thus any) $i$, we have $A_i \subseteq X$, we have $x \in X$.

(it is a theorem of set theory that: $A \subseteq B \iff [x \in A \implies x\in B$]).

Thus (see my parenthetical remark above), $\bigcup\limits_i A_i \subseteq X$, QED.

Applying this to $\langle g\rangle$, for each $g \in G$, we see that all we need to do is show that $\langle g\rangle \subseteq G$.

But since $\langle g\rangle = \{g^k: k \in \Bbb Z\}$ (if $g$ is of finite order not all of these will be distinct, but this does account for everything in $\langle g\rangle$), we see that $\langle g\rangle \subseteq G$, because $G$ is closed under its group operation.

 

[evinda]

Applying this to $\langle g\rangle$, for each $g \in G$, we see that all we need to do is show that $\langle g\rangle \subseteq G$.

But since $\langle g\rangle = \{g^k: k \in \Bbb Z\}$ (if $g$ is of finite order not all of these will be distinct, but this does account for everything in $\langle g\rangle$), we see that $\langle g\rangle \subseteq G$, because $G$ is closed under its group operation.

Could you explain to me further the following part?


we see that $\langle g\rangle \subseteq G$, because $G$ is closed under its group operation.

 

[Deveno]

Could you explain to me further the following part?


we see that $\langle g\rangle \subseteq G$, because $G$ is closed under its group operation.

By definition, a group is a set equipped with a binary operation.

By definition, a binary operation is a function $G \times G \to G$.

Thus, for every $a,b \in G$, we have $a\ast b \in $G.

In particular, if $g \in G$, then:

$g^1 = g \in G$.

and if $g^{n-1} \in G$, then $g\ast g^{n-1} = g^{1+(n-1)} = g^n \in G$.

So by induction on $n$, we have $g^n \in G$, for every $n \in \Bbb Z^{+}$.

Since by definition, $g^0 = e_G$, and every group contains an identity, we have $g^0 \in G$.

Since by definition, for $n > 0$, we have $g^{-n} = (g^n)^{-1}$, and groups contain inverses of every element in them, all such negative powers of $g$ are in $G$.

Thus $g^k \in G$, for any integer $k \in \Bbb Z$.

So $\langle g\rangle \subseteq G$, because every element of $\langle g\rangle$ is in $G$.

Also: $\langle g\rangle$ is THE SUBGROUP (of $G$) generated by $g$, and subgroups are subsets.