# Thread: Prove C(H) is a subgroup of G

1. ## Prove C(H) is a subgroup of G

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.

2. Originally Posted by mandy123
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.
You know the definition of what it means being a subgroup? Just check all the conditions which need to be satisfied. This should be a straightforward problem, just post what you did.

3. 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?

4. Originally Posted by mandy123
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.