If H is asubgroup of G then by the centralizer C(H) of H we mean the set { x in G | xh=hx for all h in H}?

prove that C(H) is asubgroup of G

If H is asubgroup of G then by the centralizer C(H) of H we mean the set { x in G | xh=hx for all h in H}?
C(H) is a subgroup of G if and only if C(H) is a group and all elements of C(H) are in G.





It's very easy to prove that all elements of C(H) are in G. All h in H are also in G (H is a subgroup), all x in G are in G, and by the definition of a group, xh and hx are also in G.





To prove C(H) is a group, we must prove closure, associativity, identity and inverse.





Closure: take two elements p,q from C(H). We know by definition of C(H) that p can be written as xh=hx, with x in G and h in H, and q can be written as yj, with y in G and j in H. The product of these elements, pq, can be written as pq=(hx)p=h(xp) and pq=p(yj)=(py)j. Since xp and py are elements of G and h and j are elements of H, we know that pq is an element of both GH and HG, so it's an element of C(H). C(H) is thus closed under its operation.





Associativity follows from the fact that G is associative, and all elements in C(H) are products of elements of G.





Identity: C(H) has an identity element e=gg (where g is G's identity element and therefore also H's identity element), for which ep=pe=e holds for each element p=hx=xh of C(H):





ep=(gg)(hx)=g(gh)x =g(hg)x=(gh)(gx)=(hg)(xg)


pe=(hx)(gg)=h(xg)g =h(gx)g=(hg)(xg)


(hg)(xg)=(h)(x)=hx=p





Inverse: there is an element q=mn=nm for every p=hx=xh in C(H) for which holds that pq=qp=e; m is h's inverse, n is x's inverse:





pq=(mn)(xh)=m(nx)h= meh=mhe=(mh)e=ee=e


qp=(hx)(nm)=h(xn)m= hem=ehm=e(hm)=ee=e





Since it has all properties of a group, C(H) is a group. So by the definition of a subgroup, C(H) is a subgroup of G.
Reply:this is perfect answer thank u Report It

Reply:You have to check 2 things


1 if x,y E C(H) then xy EC(H)


2 if xEC(H) then x^-1EC(H)