 so let's eventually get to the definition of the center of a group and if we have a group g and we make it some set and a binary operation on that we call the center of a group the center we write z and then the group name in parentheses there and we're actually going to show that this is a subgroup a subgroup of g and we are going to define z g the center of g that's all the elements g elements well we'll do that because it is a group so it's g element of g set such that s binary operation with g equals g binary operation with s for all s elements of g set remember then also the binary operation and close this is a group so this is a set of all these elements and it's the elements in g such that we have this commutative property with all of the elements in the set so certain of these elements will show this commutative property under this binary operation we have to show that it's a subgroup by showing that it has the four properties of a group so first of all we need closure so we need if g1 and g2 is an element of the center we must have the g1 composed with g sub 2 the binary operation must that must also be an element of the of the center and how do we do that well let's just write let's just write g1 composed of g2 composed with s well if we write this with associativity we could do this and remember g2 is an element of this so we can rewrite that as s composed with g2 so that's exactly the same thing and by associativity I can have these two together composed with g2 g1 is an element of the center so I can do commutativity there so that is going to be s composed of g1 composed with g2 and remember I started with this I started with g1 composed of g2 composed with s and that is equal to this and I can do that so this with s and s with that well that is just falls within the definition of this this new element this new element that I get by the binary operation by composing these two is commutative as far as all the elements s is just being an arbitrary element in other words by definition whatever this is must also be an element so we have closure so if g1 and g2 is in the center the binary operation between those are also there we're going to have associativity because this is a set I mean just taking it from a set the identity element is very easy because you know by definition of an identity element it does commute so that's not a problem and now we just have to show that every element g that's in there its inverse is also there so we've basically got to show the g inverse composed with s that equals s composed with g inverse so how can we how can we do that well let's let's have a let's have a little look at on this side what can we do on this side well what shall we do let's let's start rather with this side let's say g inverse composed of s composed of g composed of g inverse that equals s o g inverse now I've done nothing I'm just composing with the identity element I'm composing with the identity element but if I can do this I can do associativity g I'm starting off with g being an element of the center of g I can do that so I'm going to have g inverse composed with g composed with s because it is an element I can do that composed with g inverse composed with g inverse what I'm left with here in the end is just the identity element so I have that g os as we start g inverse composed with s equals s composed with g inverse and that once again satisfies the definition that I have here so I do have closure so in other words no problem I have closure I have associativity I have an identity element and every element in the center also has its inverse being in the center of g so we can really show that the center of a group is a subgroup of our original group