# Centralizer of a subgroup is a subgroup of the main group

• March 1st 2010, 08:10 AM
killercatfish
Centralizer of a subgroup is a subgroup of the main group
Hello! This is my first post here, and I believe I checked all previous posts and could not find one like this.

The question states:
If H is a subgroup 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 a subgroup of G.

I have attempted to use the 1 step subgroup test:

1. C(H)={x in G | xh=hx for all h in H}.
2. e is in C(H) since he=eh for all h in H, so C(H) is not empty.
3. Assume a,b is in C(H).
4. Show ab^-1 is in C(H).
a in C(H) mean for all h in H, h is in G, ah=ha
b in C(H) mean for all h in H, h is in G, bh=hb
I get stuck here, obviously with the majority of the proof to go, the main issue is how do I prove that it is a subgroup of G and not simply that it is a subgroup of H? It makes sense that if it is a subgroup of H, which is a subgroup of G, that it is, but how do I proceed?

Thanks!
• March 1st 2010, 08:16 AM
HallsofIvy
First, if ah= ha for all h, then, multiplying both sides of that equation by $a^{-1}$ on the left,
$a^{-1}(ah)= h= a^{-1}ha$.
But then, multiplying both sides of that by $a^{-1}$ on the right,
$ha^{-1}= (a^{-1}ha)(a^{-1}= (a^{-1}h$.

That is, if a is in the centralizer, then so is [tex]a^{-1}[tex]

Of course, if a and b are both in the cetralizer then (ab)h= a(bh)= a(hb)= (ah)b= (ha)b= h(ab).