Proof on Groups with Prime Order Elements

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Problem

Suppose that $G$ is a finite group with the property that every non-identity element has prime order. If the center of the group, $Z(G)$, is non-trivial, prove that every non-identity element of $G$ has the same order.

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Definitions Used

1. Prime Order: An element $a \neq e$ has prime order if its order, $o(a)$, is a prime number.
2. Center of a Group, $Z(G)$: The set of elements that commute with every element in $G$. Formally, $Z(G) = \{z \in G \mid zg = gz \text{ for all } g \in G\}$.
3. Non-trivial Center: The center $Z(G)$ contains at least one element other than the identity, $e$.
4. Key Property 1: The order of an element is equal to the order of its inverse: $o(a) = o(a^{-1})$.
5. Key Property 2: If $o(x) = n$ and $x^m = e$ for some integer $m$, then $n$ must divide $m$.

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Resolution

Since $Z(G)$ is non-trivial, we can choose an element $a \in Z(G)$ such that $a \neq e$. By the group’s property, its order must be a prime number. Let $o(a) = p$.

Now, let $b$ be any other non-identity element in $G$. Its order must also be a prime number. Let $o(b) = q$. Our goal is to prove that $p=q$.

Since $a \in Z(G)$, it commutes with all elements of $G$, so $ab = ba$. Let’s consider the element $ab$ raised to the power of $pq$:

$$ (ab)^{pq} = a^{pq} b^{pq} = (a^p)^q (b^q)^p = (e)^q (e)^p = e $$

The result $(ab)^{pq} = e$ implies that the order of the element $ab$ must divide $pq$.

According to the properties of group $G$, the order of any non-identity element must be prime. The order of $ab$ is therefore either a prime number or it is 1 (if $ab=e$). The only prime divisors of $pq$ are $p$ and $q$. Thus, the possible orders for $ab$ are $1, p,$ or $q$. We will now examine each case.

Case 1: $o(ab) = 1$

If the order is 1, then $ab = e$, which implies $a = b^{-1}$. Using the property that the order of an element is equal to the order of its inverse, we have:

$$ o(a) = o(b^{-1}) = o(b) $$ $$ \implies p = q $$

Case 2: $o(ab) = p$

If the order is $p$, then $(ab)^p = e$. Since $a$ and $b$ commute, we can write $a^p b^p = e$.

By definition, $o(a) = p$, so $a^p = e$. The equation becomes $e \cdot b^p = e$, which simplifies to $b^p = e$.

We now know that $o(b) = q$ and $b^p = e$. This requires that $q$ must divide $p$. Since both $q$ and $p$ are prime numbers, this is only possible if $q=p$.

Case 3: $o(ab) = q$

If the order is $q$, then $(ab)^q = e$. Since $a$ and $b$ commute, we have $a^q b^q = e$.

By definition, $o(b) = q$, so $b^q = e$. The equation becomes $a^q \cdot e = e$, which simplifies to $a^q = e$.

We now know that $o(a) = p$ and $a^q = e$. This requires that $p$ must divide $q$. Since both $p$ and $q$ are prime numbers, this is only possible if $p=q$.

Conclusion

In all possible cases, we arrive at the conclusion that $p=q$. Since $b$ was chosen as any arbitrary non-identity element, it follows that every non-identity element of $G$ has the same prime order.