- #1
- 1,462
- 44
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##?