Explanation of first part of proof that if G is any group Z(G) is a normal subgroup of G. Here's the proof of this fact from Schaum's book on group theory -
$1 \in Z(G)$ since $1g = g1$ for all $g \in G$. Consequently $Z(G) \neq \emptyset$. 
If $g_1,g_2 \in Z(G)$ and $g \in G$, then $g(g_1 g_2^{-1}) =(g g_1) g_2^{-1} = g_1(g g_2^{-1}) = g_1 g_2^{-1}g$ since $g g_2 = g_2 g$ implies $g_2^{-1} g = g g_2^{-1}$. It follows that $Z(G)$ is a subgroup of $G$.
How does it follow that $Z(G)$ is a subgroup of $G$? I am not sure what this proof has done here to show that $Z(G)$ is a subgroup of $G$. It seems to have demonstrated closure..but I'm not sure if that was the goal as there should be brackets on the final $g_1 g_2^{-1}g$ to make it obvious that that was what they were doing, ie write it as $(g_1 g_2^{-1})g$.
Anyway, could someone explain to me exactly what the process is that they used to show that its a subgroup. Have they implicity demonstrated that all of the group axioms hold?
 A: One can check to see that a sufficient condition for a subset $H\subseteq G$ to be a subgroup is that $1\in H$ and that $\forall g_1,g_2\in H,\:g_1g_2^{-1}\in H$:
It follows simply by the fact that for any $g\in H$, $g^{-1}=1\cdot g^{-1}\in H$ (since $1\in H$)- thus $H$ is closed under inversion, and likewise for any $g_1,g_2\in H$, $g_2^{-1}\in H$ by the previous argument, and thus
$$g_1g_2=g_1(g_2^{-1})^{-1}\in H$$
and consequently $H$ is closed under the group operation. All other group axioms are inherited directly from those of $G$. 
What the authors did in the segment you quoted was to show that if $g_1\in Z(G)$ and $g_2\in Z(G)$ then $g_1g_2^{-1}\in Z(G)$, and since $1$ is always in $Z(G)$ the fact that $Z(G)$ is a subgroup of $G$ follows immediately
A: As you noted, with parentheses added, it would have been easier recognized that he has shown


*

*$Z(G)\ne \emptyset$

*$g_1, g_2 \in Z(G)\Rightarrow g_1 g_2^{-1}\in Z(G)$


Indeed, these two items together show that $Z(G)$ is a subgroup.
This  "subgroup criterion" should be known; all remaining properties of a group follow from the fact that $G$ is a goup and need not be shown.
By the way, he proves and uses an intermedite result, namely $g_2\in Z(G)\Rightarrow g_2^{-1}\in Z(G)$. In fact, another form of subgroup criterion consists of these three parts:


*

*$Z(G)\ne \emptyset$

*$g_1, g_2 \in Z(G)\Rightarrow g_1 g_2\in Z(G)$

*$g_1\in Z(G)\Rightarrow g_1^{-1}\in Z(G)$


This form could thus have been used anyway (unless this form of the subgroup criterion is not used at all in the book).
