I am trying to solve this question: The center of a group $G$ (denoted by $Z$) is defined to be the set of elements $$\{z_1, z_2,...\}$$ that commute with all elements of $G$, that is $z_ig=gz_i$ for all $g \in G$. Show that $Z$ is an abelian subgroup of $G$.

So I know that if it's a subgroup of $G$ and it commutes with the elements of $G$ it must commute with itself, but I'm not sure how to prove that it is a subgroup. Don't I have to know how many elements are respectively in $Z$ and $G$ to check if Lagrange's Theorem applies?

  • 1
    $\begingroup$ No. A subgroup's order is a divisor of the group's order, but the converse is not true: a subset's size being a divisor of the group's size does not mean the subset is a subgroup. The very definition of "subgroup" is basically a list of things to check to determine if something is a subgroup. So make sure you go and look at the definition. $\endgroup$ Aug 8 '16 at 0:25

Basically you have to check the following three facts:

  • Presence of inverse elements If $zg = g z$ for all $g \in G$ you obtain by multiplication that $z^{-1}(zg)z^{-1} = z^{-1}(g z)z^{-1}$ and then $$gz^{-1} = z^{-1}g$$

    and $z^{-1}$ commutes with all $G$ and hence is contained in $Z$.

  • Presence of identity It is obvious that $e$ is contained in $Z$ because it commutes with all elements of $G$.

  • Closure by multiplication You have to show that $z \cdot z'$ is in $Z$ if $z, \, z' \in Z$, but this is easy because $$z z' g = z(z'g)= z (g z')= (zg)z'=(gz)z'=gzz'$$


The definition of a subgroup is the following:

Definition: A subset $H \subseteq G$ of a group $G$ is called a subgroup if the following condititons are satisfied:

  1. $1 \in H$, where $1$ denotes the neutral element of $G$,
  2. for all $x, y \in H$ one has $x y \in H$, and
  3. for every $x \in H$ one has $x^{-1} \in H$.

To check that $$ Z = \{z \in G \mid \text{$zg = gz$ for all $g \in G$}\} $$ is a subgroup of $G$ you need to check that this subset satisfies all three of the above conditions.

Regarding your proposed use of Lagrange's theorem:

Theorem: (Lagrange) If $H \subseteq G$ is a subgroup of a finite group $G$, then $|H|$ divides $|G|$.

First off, $G$ is not assumed to be finite, so we already cannot apply the theorem. But even if we assume $G$ to be finite, we gain nothing from this: The theorem only tells us that if we already know that $Z$ is a subgroup, then we can conclude that $|Z|$ divides $|G|$.

The converse of the theorem does not hold, i.e. if we have a subset $H \subseteq G$ such that $|H|$ divides $|G|$, then it does not necessarily follow that $H$ is a subgroup of $G$. So even if $G$ was finite and $|Z|$ would divide $|G|$, we could not use this to conclude that $Z$ is a subgroup.

The only thing we could conclude from the theorem is that if $G$ was finite and $|Z|$ would not divide $|G|$, then $Z$ would not be subgroup of $G$. But of course this won’t happen, as it is a subgroup.

  • $\begingroup$ Ohhhhh I see now thank you for the bit about Lagrange's Theorem, that really cleared things up :). $\endgroup$ Aug 8 '16 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.