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Prove that if a finite abelian group has order a power of a prime*p*, then the order of every element in the grour is a power of*p*. Can the hypothesis of commutativity be dropped? Why, or why not?

To address this problem, we are dealing with finite groups of order that is a power of a prime \( p \), and we are tasked with proving that every element in such a group has an order that is also a power of \( p \). We will also examine whether the assumption that the group is Abelian (commutative) can be dropped.

### Proof for Abelian Groups:
Let \( G \) be a finite Abelian group such that the order of \( G \) is \( p^n \), where \( p \) is a prime number and \( n \) is a non-negative integer. By Lagrange's Theorem, the order of any element \( g \in G \) divides the order of the group. Since the order of \( G \) is \( p^n \), this means that for every element \( g \in G \), the order of \( g \) (denoted \( |g| \)) must divide \( p^n \). Therefore, the order of every element must be a power of \( p \).

The key here is that the group is Abelian, which allows us to apply the classification theorem for finite Abelian groups. This theorem states that every finite Abelian group whose order is \( p^n \) can be decomposed into a direct product of cyclic groups of order powers of \( p \). Since every cyclic group of order \( p^k \) (for some \( k \)) has elements whose orders are powers of \( p \), the result follows that the order of every element in the group is a power of \( p \).

### Can the Hypothesis of Commutativity be Dropped?
The answer is **no**—the hypothesis of commutativity cannot be dropped. Without commutativity, there are examples of non-Abelian \( p \)-groups where not every element has an order that is a power of \( p \).

#### Example:
Consider the **quaternion group** \( Q_8 \), which is a non-Abelian group of order \( 8 = 2^3 \) (so, it's a \( 2 \)-group). The elements of \( Q_8 \) are:
\[
\{ 1, -1, i, -i, j, -j, k, -k \}
\]
While the orders of \( 1 \), \( -1 \), and the elements \( i, j, k \) are powers of \( 2 \), the quaternion group is non-Abelian, and not all elements satisfy the criteria for being in an Abelian structure. Therefore, the classification that works for finite Abelian groups does not apply to non-Abelian groups in general.

Thus, the hypothesis of commutativity is necessary for the proof that every element has order a power of \( p \).

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Would you like further details on specific points of the proof, or have any questions about other aspects of group theory?

Here are 5 related questions for further exploration:
1. What is the structure theorem for finitely generated Abelian groups?
2. Can you prove that the center of a \( p \)-group is non-trivial?
3. What is the role of Lagrange's theorem in understanding the order of elements in finite groups?
4. How do Sylow theorems apply to groups of prime power order?
5. How does the classification of finite simple groups connect to \( p \)-groups?

**Tip:** When working with non-Abelian groups, be cautious about assuming properties like commutativity—many key theorems for Abelian groups may not hold.

Prove that if a finite abelian group has order a power of a prime p, then the order of every element in the group is a power of p. Can the hypothesis of commutativity be dropped? Why, or why not?

Learn how to prove that if a finite Abelian group has order a power of a prime p, then the order of every element in the group is also a power of p. The solution includes the use of Lagrange's theorem and the classification of finite Abelian groups. Also, explore whether the commutativity hypothesis can be dropped and why it's necessary for the proof.

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