Central implies normal

Statement
Suppose $$H$$ is a subgroup of $$G$$ that is a central subgroup of $$G$$, i.e., $$H$$ is contained in the center $$Z(G)$$ of $$G$$. Then, $$H$$ is a normal subgroup of $$G$$.

Intermediate properties
Here's a more complete list:

Converse
The converse is not true in general: normal not implies central.

However, some versions are true:


 * Normal of least prime order implies central
 * Totally disconnected and normal in connected implies central
 * Cartan-Brauer-Hua theorem

Proof using coset definition of normality
Given: A group $$G$$, a central subgroup $$H$$ of $$G$$.

To prove: $$gH = Hg$$ for all $$g \in G$$ (the cosets definition of normality).

Proof: Since $$H$$ is central, this means that $$gh = hg$$ for all $$g \in G$$, $$h \in H$$. Thus, for a fixed $$g$$, the sets $$gH = \{ gh: h \in H \}$$ and $$Hg = \{ hg : h \in H \}$$ are equal, because each $$gh$$ equals the corresponding $$hg$$.

Proof using conjugation definition of normality
Given: A group $$G$$, a central subgroup $$H$$ of $$G$$.

To prove: For all $$g \in G$$ and $$h \in H$$, we have $$ghg^{-1} \in H$$.

Proof: Since $$h$$ is central, we have, by definition, that $$gh = hg$$ for all $$g \in G$$. Multiplying both sides on the right by $$g^{-1}$$, we obtain that $$ghg^{-1} = h$$ for all $$g \in G$$. Since $$h \in H$$ by assumption, and $$ghg^{-1} = h$$, we obtain that $$ghg^{-1} = h$$.

Proof using commutator definition of normality
Given: A group $$G$$, a central subgroup $$H$$ of $$G$$.

To prove: For all $$g \in G$$, $$h \in H$$, we have $$ghg^{-1}h^{-1} \in H$$.

Proof: Since $$h \in H$$ we have $$gh = hg$$ by definition. Multiplying both sides by $$g^{-1}h^{-1}$$ on the right, we get $$ghg^{-1}h^{-1} = e$$ (i.e., it is the identity element). Since any subgroup contains the identity element, $$e \in H$$, so $$ghg^{-1}{h^{-1} \in H$$.

Proof using union of conjugacy classes definition of normality
Given: A group $$G$$, a central subgroup $$H$$ of $$G$$.

To prove: $$H$$ is a union of conjugacy classes in $$G$$.

Proof: Every element of the center of $$G$$ forms a conjugacy class of size 1. Since $$H$$ comprises only central elements, it is the union of these singleton conjugacy classes.