I Generalizing the definition of a subgroup

1. Feb 20, 2017

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 $a$?

2. Feb 21, 2017

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 $G$.