Final answer:
The set Ha = x ∈ G for a fixed element a in a group G contains all elements of G that commute with a and is a subset of G, making Ha a legitimate set in the context of group theory.
Step-by-step explanation:
Show that Ha={x∈G∩xa=ax} is a set in group G
In group theory, a branch of mathematics, we're often interested in sets that have particular properties related to the operation of the group. Let's consider the set Ha defined as Ha={x∈G∩xa=ax} for a fixed element a in group G. Our task is to explore the properties of this set.
Properties of Ha
We need to show that for every element x in group G, if xa equals ax, then x is an element of the set Ha. To prove that Ha is a subset of G, we can consider two cases:
- For any element x in Ha, by definition of the set, xa=ax. So every element in Ha satisfies the condition of commutativity with the fixed element a. Hence, every element x in Ha is also an element of G.
- If an element y is not in Ha, then ya ≠ ay. This means that the element y does not satisfy the commutativity with a and is not included in the set Ha.
The set Ha is therefore clearly defined within the group G since it contains elements of G that commute specifically with the element a. The elements of Ha are a subset of the elements of G by definition, making Ha a legitimate set associated with group G.