Generalizing the definition of a subgroup

[Mr Davis 97]

Let ##G## be a group. I have shown that ##H_a = \{x \in G | xa=ax \}## is a subgroup of G, where ##a## is one fixed element of ##G##. I am now asked to show that ##H_S = \{x \in G ~| ~xs=sx,~ \forall s \in S\}## is a subgroup of ##G##. How would proving the former differ from proving the latter? Couldn't I essentially use the same proof as the former but use ##s## instead of ##a##?

 

[andrewkirk]

The proof is not the same, because the elements of ##H_a## only need to commute with the single element ##a##, whereas the elements of ##H_S## needs to commute with all elements of ##S##, which we assume (although it is not stated) is a subset (not necessarily a subgroup) of ##G##. If ##S## has only one element, the proofs will be the same. Otherwise not, although they may be similar.

By the way, if ##S=G## then ##H_S## is called the 'centre' of the group ##G##.