Hey I am having trouble formalizing part (c).



I get that I am going backwards from my proof in part (a) but I'm getting hung up on things like how the identity element is in G but how do I say its in S?

Similarly how do I prove existence of inverses? I think I use symmetry but can't really flesh it out. Same with using transitivity to show associativity.

And lastly how do I show closure?

Thanks!

First let's show existence of identity
By the equivalence relation on G we can have
a~a => aa^-1 ∈ S for a ∈ G
but aa^-1 is the identity element e
so e ∈ S

Next show inverse
for a,b ∈ G by symmetry
a~b and b~a => ab^-1 and ba^-1
(ab^-1)(ab^-1)^-1 ∈ S
this is because
(ab^-1)(ab^-1)^-1 = (ab^-1)ba^-1
from associativity in G we get
=a(b^-1b)a^-1
=e
And we know e ∈ S, so S is closed under inverses

Associative
This is easy, since the operation for S is the same as the operation on G, the operation is associative since G is a group.

Closure
Suppose a,b ∈ S. We showed S is closed under inverses, so we know b^-1 ∈ S
then a~b^-1 => a(b^-1)^-1 = ab ∈ S

In case you were wondering why the bolded part is true.
a~e, and b^-1 ~ e and by symmetry e~b^-1
Finally using transitive property a~b^-1

thank you!