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**tttcomrader** Problem: Let G be a group, prove that the centralizers of G is a subgroup of G.

Proof: By definitions, the centralizers of G, $\displaystyle C(a) = \{ g \in G:ga=ag \} \forall a \in G$

Now, the identity element, e, has the property of $\displaystyle ea=ae$, thus e is in C(a). So C(a) is not an empty set.

Assume C(a) contains more than just {e}, since if that is the case C(a) would be a subgroup.

Let x,y be in C(a), then: $\displaystyle xa=ax,ya=ay \forall a \in G$

Then $\displaystyle (xy)a=x(ya)=x(ay)=(xa)y=a(xy)$

Thus xy is in C(a).

Now, $\displaystyle (xa)^{-1}=(ax)^{-1}$

$\displaystyle x^{-1}a^{-1}=a^{-1}x^{-1}$

So x^{-1} is in C(a), thus C(a) is a subgroup of G.

Q.E.D.