Tour:Nonempty finite subsemigroup of group is subgroup: Difference between revisions

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This article adapts material from the main article: subsemigroup of finite group is subgroup

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In a finite group, any nonempty subset that is closed under the operation of multiplication is, in fact, a subgroup. This isn't true for infinite groups (think for a moment about positive integers inside all integers).
WHAT YOU NEED TO DO: Understand the statement and proof below.

Statement

Verbal statement

Any nonempty multiplicatively closed finite subset (or equivalently, nonempty finite subsemigroup) of a group is a subgroup.

Symbolic statement

Let $G$ be a group and $H$ be a nonempty finite subset such that $a, b \in H ⟹ a b \in H$ . Then, $H$ is a subgroup of $G$ .

Proof

Lemma

Statement of lemma: For any $x \in H$ :

All the positive powers of $x$ are in $H$
There exists a positive integer $n (x)$ , dependent on $x$ , such that $x^{n (x)} = e$ .

Proof: $H$ is closed under multiplication, so we get that the positive power of $x$ are all in $H$ . This proves (1).

Since $H$ is finite, the sequence $x, x^{2}, x^{3}, \dots$ must have some repeated element. Thus, there are positive integers $k > l$ such that $x^{k} = x^{l}$ . Multiplying both sides by $x^{- l}$ , we get $x^{k - l} = e$ . Set $n (x) = k - l$ , and we get $x^{n (x)} = e$ . Since $k > l$ , $n (x)$ is a positive integer.

The proof

We prove that $H$ satisfies the three conditions for being a subgroup, i.e., it is closed under all the group operations:

Binary operation: Closure under the binary operation is already given to us.
Identity element $e \in H$ : Since $H$ is nonempty, there exists some element $u \in H$ . Set $x = u$ in the lemma. Applying part (2) of the lemma, we get that $e$ is a positive power of $u$ , so by part (1) of the lemma, $e \in H$ .
Inverses g∈H⟹g−1∈H: Set x=g in the lemma. We make two cases:
- Case $n (g) = 1$ : In this case $g = e$ forcing $g^{- 1} = e = g \in H$ .
- Case $n (g) > 1$ : In this case $g^{- 1} = g^{n (g) - 1}$ is a positive power of $g$ , hence by Part (1) of the lemma, $g^{- 1} \in H$ .

Related results

Sufficiency of subgroup condition

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 __NOTOC__
 {{derivative of|subsemigroup of finite group is subgroup}}
-{{guided tour|beginners|Introduction two|Sufficiency of subgroup criterion|Finite group}}
+{{tour-regular|
-{{quotation|In a finite group, any nonempty subset that is closed under the operation of multiplication is, in fact, a subgroup. This isn't true for infinite groups (think for a moment about positive integers inside all integers). See below for the proof and more details.<br>There is a ''general'' condition called the subgroup condition for arbitrary groups, as we'll see in the next article.<br>Proceed to [[Guided tour for beginners:Sufficiency of subgroup criterion]] OR return to [[Guided tour for beginners:Finite group]]}}
+target = beginners|
-==Statement==
+secnum = two|
+next = Sufficiency of subgroup criterion|
+previous = Finite group|
+expected time = 8 minutes}}
+{{quotation|In a finite group, any nonempty subset that is closed under the operation of multiplication is, in fact, a subgroup. This isn't true for infinite groups (think for a moment about positive integers inside all integers).<br>'''WHAT YOU NEED TO DO''': Understand the statement and proof below.}}
-===Verbal statement===
+{{#lst:Subsemigroup of finite group is subgroup|main}}
-Any nonempty multiplicatively closed subset (or equivalently, nonempty subsemigroup) of a [[finite group]] is a [[subgroup]].
+{{tour-regular-bottom|
+target = beginners|
-===Symbolic statement===
+secnum = two|
+next = Sufficiency of subgroup criterion|
-Let <math>G</math> be a [[finite group]] and <math>H</math> be a nonempty subset such that <math>x,y \in H \implies xy \in H</math>. Then, <math>H</math> is a subgroup of <math>G</math>.
+previous = Finite group}}
-==Proof==
-===Finite order of every element===
-We use finiteness to observe that <math>x</math> has finite [[order of an element|order]], or equivalently, for any <math>x \in G</math> that there exists a <math>n</math> such that <math>x^n = e</math>. This can be shown as follows:
-For any <math>x \in G</math>, the set <math>x,x^2,x^3, \ldots</math> is a finite set and hence there are positive integers <math>k</math> and <math>l</math> such that <math>x^k = x^l</math>. This gives <math>x^{k-l} = e</math>.
-===The proof===
-Since <math>H</math> is nonempty, there exists <math>x \in H</math>. We now demonstrate the three conditions for <math>H</math> to be a subgroup:
-* '''Binary operation''': The product of two elements in <math>H</math> is in <math>H</math>, by assumption
-* '''Identity element''': There exists <math>x \in H</math> and from the previous section we know that there exists <math>n</math> such that <math>x^n = e</math>. Since <math>H</math> is closed under multiplication, <math>e \in H</math>
-* '''Inverse element''': For any <math>x \in H</math>, there exists <math>n</math> such that <math>x^n = e</math>. Hence, <math>x^{n-1} = x^{-1}</math>. Since <math>H</math> is closed under multiplication, <math>x^{n-1} \in H</math> and hence <math>x^{-1} \in H</math>.
-==Related results==
-* [[Sufficiency of subgroup condition]]
-==External links==
-===Links to related riders===
-{{mathlinks|134285|Subgroup of a finite group}}