Question

Let G be a group and $\mathit{a}\mathbf{\in }\mathit{G}$. The centralizer of a is the set$\mathit{C}\left(a\right)\mathbf{=}\left\{g\in G|ga=ag\right\}$. Prove that$\mathit{C}\left(a\right)$ is a subgroup of G.

Short answer

Itis proved that C(a) is a subgroup of G.

Step-by-step solution

Consider the given information

Consider an element $\mathit{a}$of group $\mathit{G}$. Then, the centralizer of $\mathit{a}$the element is a set defined as$\mathit{C}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{g}\mathbf{\in }\mathbf{G}\mathbf{|}\mathbf{ga}\mathbf{=}\mathbf{ag}\mathbf{\right\}}$.

Prove C(a)that is a subgroup of G

Consider two elements and in the set with the following properties:

$pa=ap$

And,

$qa=aq$

Now, consider the product of these two elements $pq$.

Perform the following operation:

$$\begin{gathered} pqa=p\left(qa\right) \\ =p\left(aq\right) \\ =\left(pa\right)q \\ =apq \end{gathered}$$

Therefore, the element $pq$is also present in set $C\left(a\right)$.

Consider two elements $p$in set $C\left(a\right)$with the following property;

$pa=ap$

Perform the following operation.

$$\begin{gathered} p^{-1}\left(pa\right)=p^{-1}\left(ap\right) \\ ea=p^{-1}ap \\ a{p}^{-1}=p^{-1}ap{p}^{-1} \\ a{p}^{-1}=p^{-1}a \end{gathered}$$

Therefore, element $p$is also present in set$C\left(a\right)$. So, the centralizer is a subgroup of$G$.

Hence, it is proved that$C\left(a\right)$ is a subgroup of G.