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.


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.

  • 1
    $\begingroup$ There is not much to do - the difference is on which side of the such that $a$ lies. $\endgroup$ – 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: $\displaystyle\bigcap_{a \in G} C(a) = Z(G)$.

So, if the centralizer of a particular element is the whole group, that means every element commutes with it, so it must be in the center. Conversely, if it's in the center, then every element commutes with it, so every element is in its centralizer.

  • $\begingroup$ Wow, excellent explanation. I understood it better then in lecture. Despite we didnt talk about the center itself a lot, only about centrelizers, but that makes some order in that definitions. Thank you! $\endgroup$ – Ilya.K. Aug 14 '15 at 19:27

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.


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