# Tour:Nonempty finite subsemigroup of 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  be a group and  be a nonempty finite subset such that . Then,  is a subgroup of .

## Proof

### Lemma

Statement of lemma: For any :

1. All the positive powers of  are in 
2. There exists a positive integer , dependent on , such that .

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

Since  is finite, the sequence  must have some repeated element. Thus, there are positive integers  such that . Multiplying both sides by , we get . Set , and we get . Since ,  is a positive integer.

### The proof

We prove that  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 : Since  is nonempty, there exists some element . Set  in the lemma. Applying part (2) of the lemma, we get that  is a positive power of , so by part (1) of the lemma, .
• Inverses : Set  in the lemma. We make two cases:
• Case : In this case  forcing .
• Case : In this case  is a positive power of , hence by Part (1) of the lemma, .