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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?

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    $\begingroup$ 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. $\endgroup$ – arctic tern Aug 8 '16 at 0:25
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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.

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  • $\begingroup$ Ohhhhh I see now thank you for the bit about Lagrange's Theorem, that really cleared things up :). $\endgroup$ – Shrodinger 2016 Aug 8 '16 at 2:38
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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 $$z z' g = z(z'g)= z (g z')= (zg)z'=(gz)z'=gzz'$$

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