# The difference between Z(G) and C(a) in an example

I found that I didnt understood the defenitions. I have this exercize: to prove that $a\in Z(G)$ $<==>$ $C(a) = G$

Is there here something to prove? Isnt it directly of their defenitions ? I hope you will understand the symbols, they are the standard for center.

Edit:

Ok, so after understanding that, lets try an example:

How do I find $Z(G)$ where $$G=\left\{T_{a,b}\:;\:a\ne 0,\:b\in R,\:T_{a,b}:\:R\:\:\vec{\:}R\:,\:T_{_{a,b}}=ax+b\right\}$$ Or by words the linear composition function.

So I dont studied it at yet and I just will try to follow my logic by what we said below.

Firstly, I want to see at least one centralizer, just for an example of how does it looks and what its characteristics, so I will had a chance to do some cocnlusion about any function in G (as I understand that is the Z(G), center as we define it). Is this right algorithm ?

so, lets look what are the conditons on $S_{c,d}\left(x\right)$ so it will be comutative with specific $T_{a,b}\left(x\right)$:

So I'm looking for: $$S_{c,d}\left(T_{_{a,b}}\left(x\right)\right)=T_{a,b}\left(S_{_{c,d}}\left(x\right)\right)$$ to find $C\left(T_{a,b\left(x\right)}\right)$, is this correct ?

after calculating the both sides of the equations, I have got this condition: $$d(a-1)=b(c-1)$$

Am I in the right direction? Now, there isnt too much diffrence to any other function instead of T, so the center is all the functions that, that condition is true, am I correct?

Otherwise, dont judge my thinking way too hardly, it is my first attempt without any special preparation. Please show me the right path to solve that kind of exercises.

• There is not much to do - the difference is on which side of the such that $a$ lies. – Michael Burr Aug 14 '15 at 19:19

$C(a)$ is called the centralizer for $a$ and is defined to be the set $\{g \in G | ga = ag\}$ for some particular $a \in G$, whereas $Z(G)$ is the set of elements that commutes with any element of $G$.
In particular, we have the identity: $\displaystyle\bigcap_{a \in G} C(a) = Z(G)$.
For the second part of your question, from what I understand you've defined $G$ to be: $$G = \{ax+b\ |\ a,b\in\mathbb{R},\ a\neq0\}$$ under function composition. The center of $G$ will be: $$Z(G) = \{f = cx+d\ |\ f\circ g = g\circ f,\ \forall g\in G\}.$$ Since the elements of $Z(G)$ need to commute with every element of $G$, we start with an arbitrary element $ax+b\in G$, and see what the conditions are on another element $cx+d\in G$ to ensure the two commute. We need: $$a(cx+d)+b = c(ax+b)+d,$$ which simplifies to: $$acx+ad+b = acx+cb+d\quad\to\quad d(a-1)=b(c-1).$$ So you were in the right direction! This tells us that in order for $cx+d$ to commute with every element of $G$, we need $c,d\in\mathbb{R},\ c\neq 0$ to obey: $$d(a-1)=b(c-1),$$ for ALL $a,b\in\mathbb{R},\ a\neq 0$. If $c,d$ do in fact obey this requirement, then they obey the requirement when $a=1$ and $b\neq 0$, which implies $c-1=0$, and so we must have $c=1$. Finally, our requirement has reduced to $$d(a-1)=0$$ for all $a\in\mathbb{R}, a\neq 0$, and so when $a=2$, this forces $d=0$. So the center is: $$Z(G) = \{cx+d\ |\ c=1,\ d=0\} = \{x\}.$$ As for the centralizer, you've already computed it for a general element of $G$: $$C(ax+b) = \{cx+d\ |\ d(a-1)=b(c-1)\},$$ which may just be $\{x\}$ again, but I'll leave that for you to play with.