Existence of an Element of Prime Order

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Problem

Let $G$ be a finite group with more than one element. Show that $G$ has an element of prime order.

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Definitions Used

1. Finite Group: A group with a finite number of elements.
2. Order of an Element, $o(a)$: The smallest positive integer $n$ such that $a^n = e$, where $e$ is the identity.
3. Prime Number: A positive integer greater than 1 that has no positive divisors other than 1 and itself.
4. Composite Number: A positive integer greater than 1 that is not prime.

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Proof

Let $G$ be a finite group with $|G| > 1$. Since the group has more than one element, we can choose an element $a \in G$ such that $a \neq e$.

Because $G$ is finite, the element $a$ must have a finite order. Let $o(a) = n$. Since $a \neq e$, we know that $n > 1$.

The integer $n$ must be either a prime number or a composite number. We consider both possibilities.

Case 1: The order $n$ is a prime number.

If $n$ is prime, then we have found an element, $a$, whose order is prime. The proof is complete for this case.

Case 2: The order $n$ is a composite number.

If $n$ is composite, then by definition, it has a prime divisor. Let $p$ be a prime number that divides $n$. We can write:

$$ n = pq, \quad \text{for some integer } q \text{ where } 1 < q < n. $$

Now, we construct a new element $b$ from $a$:

$$ b = a^q $$

Let’s determine the order of this new element $b$. We can start by raising $b$ to the power of $p$:

$$ b^p = (a^q)^p = a^{qp} = a^n $$

Since the order of $a$ is $n$, we know that $a^n = e$. Therefore:

$$ b^p = e $$

This result tells us that the order of $b$ must divide $p$. The only positive divisors of a prime number $p$ are $1$ and $p$. So, the order of $b$ must be either $1$ or $p$.

Could the order of $b$ be 1? If $o(b) = 1$, then $b=e$. This implies $a^q = e$. However, we know that $q < n$, and $n$ is defined as the smallest positive integer for which $a^n = e$. This is a contradiction.

Since the order of $b$ cannot be 1, the only remaining possibility is that the order of $b$ is $p$.

Conclusion

We have shown that in either case, we can find an element of prime order. If we pick a non-identity element and its order is prime, we are done. If its order is composite, we can use it to construct another element that is guaranteed to have prime order. Therefore, every finite group with more than one element has an element of prime order.