The discussion revolves around proving that the cyclic subgroup generated by an element \( a \) is a subset of the centralizer \( C(a) \) in the context of abelian groups. Participants are exploring the implications of the commutativity condition \( xa = ax \) and its relation to the structure of the group.
The discussion is ongoing, with various interpretations of the definitions and properties of the groups being explored. Some participants have offered clarifications about the nature of \( C(a) \) and the cyclic subgroup, while others are still grappling with the implications of their reasoning.
There is a mention of the potential confusion arising from the notation used for group operations, as well as the distinction between commutative and non-commutative groups. Participants are also considering the implications of whether \( a \) is in the center of the group.
No, it doesn't. It tells you that C(a) is the set of all members of G that commute with a. If G is commutative, then C(a)= G. Otherwise C(a) is a subset of G.kathrynag said:Homework Statement
Show <a>[tex]\subseteq[/tex]C(a) where C(a) is such that xa=ax.
Homework Equations
The Attempt at a Solution
Knowing xa=ax tells me we have an abelian group.
If you meant "proper subset", you have to further assume that a is not in the center of G.HallsofIvy said:No, it doesn't. It tells you that C(a) is the set of all members of G that commute with a. If G is commutative, then C(a)= G. Otherwise C(a) is a subset of G.