Question

Let \(G\) be a non-Abelian group with \(|G|=2 p\), where \(p\) is prime. Show that there exists a \(g \in G\) such that \(|g|=p\).

Short answer

There exists an element \(g\) in \(G\) such that \(|g| = p\) because there are 2 Sylow-\(p\) subgroups in \(G\).

Step-by-step solution

Apply Sylow's Theorems

We begin by identifying the number of Sylow- subsets, denoted as \(n_p\). According to Sylow's Theorems, \(n_p\) must satisfy \(n_p \equiv 1 \mod p\) and \(n_p\) divides the group order \(2p\). Since \(n_p\) divides \(2p\) and is also \(1 \mod p\), the possibilities for \(n_p\) are \(1\) or \(2\).

Consider the Possibility of n_p = 1

Suppose \(n_p = 1\). This implies there is a unique Sylow-\(p\) subgroup of \(G\), which must be normal. A normal Sylow-\(p\) subgroup in \(G\) suggests that all of its left cosets and right cosets coincide. This might force \(G\) to become an Abelian group, contradicting the initial condition that \(G\) is non-Abelian. Hence, \(n_p = 1\) leads to a contradiction.

Conclude n_p = 2 for the Sylow-p Subgroups

Since \(n_p\) cannot be \(1\), it must be \(2\). This implies there are exactly two distinct Sylow-\(p\) subgroups, each of order \(p\), in the group \(G\).

Analyze Existence of Elements of Order p

Within each of these two Sylow-\(p\) subgroups, there are elements of order \(p\). This follows from the fact that a subgroup of prime order \(p\) is cyclic and generated by any of its non-identity elements. Therefore, there exists some element \(g \in G\) such that \(|g| = p\).

Key concepts

Non-Abelian Group

In the world of group theory, a "non-Abelian group" is one in which at least two elements do not commute. In simple terms, if you take two elements, say \(a\) and \(b\) from the group, and \(ab eq ba\), then that's a non-Abelian group. This differing order of multiplication is what distinguishes non-Abelian groups from Abelian ones. Non-Abelian groups allow for more complex structures and their properties are fundamental in many areas of mathematics.

Key points include:

• Not all elements commute: \(ab eq ba\) for some \(a, b \in G\).
• They lead to richer interactions compared to Abelian groups, where every pair of elements commutes.
• Many symmetry and rotation groups encountered in physical applications are non-Abelian.

Group Order

The "group order" refers to the total number of elements in a group, written as \(|G|\). In this context, the group holds significant information about its structure and the types of subgroups it can have. The exercise we are looking at involves a group of order \(2p\), where \(p\) is a prime number. This total count is pivotal when applying Sylow Theorems to determine the number and type of subgroups.

Important facts about group order include:

• It is fundamental in characterizing the group's overall structure.
• Group order influences possible subgroup orders, following Lagrange's Theorem.
• Prime factors of the order give clues for applying Sylow's theorems.

Recognizing and understanding the group order enables mathematicians to explore deeper properties like the possibility of non-Abelian behavior or the existence of elements of specific order within the group.

Prime Order

When a group or subgroup is said to have a "prime order", it means the number of its elements is a prime number, like \(2, 3, 5,\) or \(7\). Such groups are always cyclic, as shown in the proof steps, and a single element can generate the entire group by repeated operations.

• A subgroup of prime order \(p\) has only the identity element and \(p-1\) elements of period \(p\).
• Understanding the prime order helps in identifying possible cyclic subgroups and their generators.

Recall, in our exercise, each Sylow- subgroup comprising prime order is inherently cyclic and proves crucial in showing that there is at least one element \(g\) in \(G\) satisfying \(|g| = p\).
Having prime order subgroups contributes greatly to determining group structure, and lays groundwork for deeper exploration using the Sylow theorems.

Cyclic Subgroup

A "cyclic subgroup" is one which can be generated by repeatedly applying the group operation to a single element, known as the generator. When you have a subgroup where every element can be expressed as some power of a single element, you have a cyclic subgroup. For example, if \(a\) is an element in a subgroup \(H\), and \(H = \{e, a, a^2, \ldots, a^{n-1}\}\), then \(H\) is cyclic.

• Cyclic subgroups are simple in structure and very predictable.
• Any group of prime order is cyclic; this is due to Lagrange's Theorem and the nature of prime numbers.
• In our exercise, knowing that subgroups of order \(p\) are cyclic helps in concluding there are elements with an order \(p\).

Cyclic subgroups provide a foundation upon which more complex group structures can be understood and analyzed.