0

Definition for the center of a Group:

The center $Z(G)$ of a group $G$ is the subset of elements in $G$ that commute with every other element of $G$.

Theorem: The center of a group $G$ is a subgroup of $G$.

The author begins the proof by stating the it is obvious that

$$e \in Z(G)$$ (why? I understand that the identity element is necessarily in the group G but is it necessarily in a subset?)

The author proceeds then to state

$$(ab)x = a(bx) = (ax)b = (xa)b = x(ab) \qquad\forall x \in G$$ and therefore $$ab \in G$$

How does the commutativity property convince the reader that $ab \in G$?

user35603
  • 3,022
  • 4
  • 24
  • 32
Mathematicing
  • 6,541
  • For your first question. $Z(G) = {a\in G: \forall b \in Gab = ba}$. So, take a look at $e$. You have $eb = b = be, \forall b \in G \implies e \in Z(G)$. – quapka May 24 '15 at 09:36
  • It is obvious that $e \in Z(G)$ since $e$ is commutative with every element of $G$ – MathematicianByMistake May 24 '15 at 09:37
  • Also, $e$ belongs to every subgroup of a group $G$,since the subgroup is..you guessed it..a group! – MathematicianByMistake May 24 '15 at 09:41
  • 1
    It seems to me, that there is a typo. When you say "and therefore $ab\in G$, shouldn't it be "and therefore $ab\in Z(G)$"? The equalities above those words do not prove, that $ab \in G$. The fact $ab \in G$ for $a,b\in G$ is because $G$ is a group, so it must satisfy the condition, that it is closed under the operation $\Leftrightarrow a,b \in G \implies ab\in G$. – quapka May 24 '15 at 09:52
  • @JDrinas but the question was about showing that $Z(G)$ is a subgroup, so one cannot use that fact here. – Marc van Leeuwen May 24 '15 at 09:52
  • You should cite the proof precisely. As it stands, there are no assumption made on $a,b$, and the sequence of equalities is therefore largely unjustified (only the first and the last equalities are true without any assumptions). Also the conclusion $ab\in G$ has no relation to the center $Z(G)$. – Marc van Leeuwen May 24 '15 at 09:56
  • @Marc van Leeuwen You are right Marc,I think the question was originally whether $e$ belongs to every subgroup-might have been edited or I was a bit hasty.Still,it holds that $e$ is commutative with every element of the group,thus it belongs to $Z(G)$.. – MathematicianByMistake May 24 '15 at 09:56

2 Answers2

3

Update

I forget about the inverses, so please check Walter's answer (I could write it also, but I think there is no need).

For your first question. $Z(G) = \{a\in G: \forall b \in Gab = ba\}$. So, take a look at $e$. You have $eb = b = be, \forall b \in G \implies e \in Z(G)$.

For the second question. We need to use the fact, that the group operation is associative ($a(bc) = (ab)c, \forall a,b,c \in G$), and that $a, b \in Z(G)$.

To prove, that $Z(G)$ is a subgroup, we need $e \in Z(G)$ (already done) and that it is closed under the operation, and closed under taking inverse(!), that means $$ \forall a, b \in Z(G) \implies ab \in Z(G) \Leftrightarrow (ab)x = x(ab), \forall x \in G. $$

Lets do it. Since $a,b \in Z(G) \implies ax=xa, bx=xb, \forall x \in G$ and we calculate $$ (ab)x\stackrel{\text{assoc.}}=a(bx)\stackrel{b\in Z(G)}=a(xb)\stackrel{\text{assoc.}}=(ax)b\stackrel{a\in Z(G)}= (xa)b\stackrel{\text{assoc.}}= x(ab). $$ And that's exactly, what we wanted. The last row of equalities means, that $ ab \in Z(G)$ whenever $a, b \in Z(G)$ (so $Z(G)$ is closed under the group operation).

quapka
  • 1,476
  • 9
  • 22
  • 1
    Looking at the definition of Z(G), suddenly it makes sense that the identity element, e, is in the center of the group since 'e' commutes with each and every other element in the group G! – Mathematicing May 24 '15 at 10:16
2

The line $(ab)x = a(bx) = (ax)b = (xa)b = x(ab), ∀x$ in $G$ implies that $ab$ commutes with arbitrary members of $G$ and so $ab \in Z(G)$. [We already know $ab \in G$ because $G$ is a group; you may have misread this part of the proof].

To prove $Z(G)$ is a subgroup, you need to show it is closed under the operation and that it is closed under taking of inverses. You've done the first one. Now to show it's closed under taking inverses. Let $a \in Z(G)$. $\forall x \in G,a^{-1}x = (x^{-1}a)^{-1} = (ax^{-1})^{-1} = xa^{-1}$. So $a^{-1} \in Z(G)$, and you're done.

Stanley
  • 3,244