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 $ab \in G$?

  • $\begingroup$ For your first question. $Z(G) = \{a\in G: \forall b \in Gab = ba\}$. So, take a look at $e$. You have $eb = b = be, \forall b \in G \implies e \in Z(G)$. $\endgroup$
    – quapka
    May 24, 2015 at 9:36
  • $\begingroup$ It is obvious that $e \in Z(G)$ since $e$ is commutative with every element of $G$ $\endgroup$ May 24, 2015 at 9:37
  • $\begingroup$ Also, $e$ belongs to every subgroup of a group $G$,since the subgroup is..you guessed it..a group! $\endgroup$ May 24, 2015 at 9:41
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    $\begingroup$ 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 $\Leftrightarrow a,b \in G \implies ab\in G$. $\endgroup$
    – quapka
    May 24, 2015 at 9:52
  • $\begingroup$ @JDrinas but the question was about showing that $Z(G)$ is a subgroup, so one cannot use that fact here. $\endgroup$ May 24, 2015 at 9:52

2 Answers 2



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).

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    $\begingroup$ 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! $\endgroup$ May 24, 2015 at 10:16

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.


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