Q:

Show that $N(a)= \{ x \in G | x\circ a = a \circ x \}$ is a sub-group when $(G,\circ)$ is a group.

So, it is given that $(G,\circ)$ is a group and we have to
show that $N(a)= \{ x \in G | x\circ a = a \circ x \}$ is a sub-group when $(G,\circ)$ is a group, and $a$ is a fixed element of a group.
I am completely blank at this one.I know what properties I need to prove an algebraic structure a group, but how do I apply them here.
Please help me in moving forward.

A:

The standard notation is $C(a)$, not $N(a)$. The centralizer of $a$ in $G$ is
$$ C(a) = \{ g\in G\mid ga=ag\}.$$
I will use that notation.
Well, to show that $C(a)$ is a subgroup of $G$, you need to show that:

$e\in C(a)$. Is it the case that $ae=ea$?
If $x\in C(a)$, then $x^{-1}\in C(a)$. Is it true that if $xa=ax$, then it follows that $x^{-1}a=ax^{-1}$?
If $x\in C(a)$ and $y\in C(a)$, then $xy\in C(a)$. If $xa=ax$ and $ya=ay$, does it follow that $(xy)a = a(xy)$?

So, check that.

