If H is a subgroup of G then the centralizer C(H) of H we mean the set C(H)={x is a member of G/ xh=hx for all h as a member of H}. Prove that C(H) is a subgroup of G.
so I think i got it, but I am not sure!
Let x be a member of G
b=x and a=x
ab^-1=xx^-1 which is a member of G, so e is a member of G
If a=e then b is a member of G
eb^-1 is a member of G so b^-1 is a member of H
x,y are members of G
x=a and y=b^-1
b=y^-1
b is a member of G so ab^-1=xy which is a member of G
Will this work or am i missing some steps?
I am not really sure what you are doing. Here, watch this, look at how we prove closure property. Let $\displaystyle a,b\in C(H)$ then it means $\displaystyle ax = xa$ and $\displaystyle bx=xb$ for all $\displaystyle x\in H$. Therefore, $\displaystyle (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab)$. Thus, $\displaystyle ab\in C(H)$. And so it is closed.