Am I using the terminology of group theory correctly? I'm attempting to learn some group theory. Is the following correct (both in terms of terminology and the validity of the conclusions written in bold)?

Define the abelian group (which is the symmetry group of some geometric object)
  $$(G,+)=\{0,x,y,z,x+y,x+z,y+z,x+y+z\},$$
  where $x,y,z$ are reflections and thus of order $2$. Because of their order, the identity, $0,$ is any of $x+x,y+y,z+z.$ For these reasons, $\bf{\{x,y,z\}}$ is the generating set of $G.$ Because $G$ is abelian and of order $8,$ $G$ is isomorphic to either $\mathbb{Z}_8,$ $\mathbb{Z}_4\oplus\mathbb{Z}_2$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2,$ where $\oplus$ denotes the direct sum. But since $G=(0,x)\oplus(0,y)\oplus(0,z),$ $\bf{G}$ must be isomorphic to the latter of the three. 

That is, am I using the concepts/terminology correctly and are the conclusions written in bold correct given the definition of $G$? Furthermore, my second conclusion seems very vague (it is not clear in my head either) - if it is true, how can it be made more rigorous?
Thanks. 
 A: Well, if we want to be rigorous then you can't define an element of $G$ as $x+y$. That's because $+$ is a function on $G$, so you need to have $G$ first. You have a circular reference here. So normally you would do 
$$G=\{0,x,y,z,a_1,a_2,a_3, a_4\}$$
and then you define $+$ as
$$x+y=a_1$$
$$x+z=a_2$$
$$y+z=a_3$$
$$...$$
But your definition can obviously be understood as a shortcut (even though with such approach it might be harder to prove that $G$ is a group).

Now for the second part you use $\oplus$ symbol with two different meanings. So let's be rigorous here as well. $G \oplus H$ is a direct sum of groups $G, H$, defined as the set of pairs $(g, h)\in G\times H$ together with pointwise group multiplication/addition.
Also if $G$ is an abelian group and $A_1,\ldots, A_n\leq G$ then define
$$A_1+\cdots+A_n=\{a_1+\cdots+a_n\ |\ a_i\in A_i\}$$
It follows that the set above is a subgroup of $G$. Now we have the following lemma

Let $G$ be an abelian group and $A_1,\ldots,A_n\leq G$. If $G=A_1+\cdots +A_n$ and for any $i$ we have $$A_i\cap (A_1+\cdots+A_{i-1}+A_{i+1}+\cdots+A_n)=0$$ then $G\simeq A_1\oplus\cdots\oplus A_n$.

Note that in one place I use equality (of sets) $=$ symbol while in the other isomorphism (of groups) $\simeq$ symbol.
This explains why $G=\{0,x\}+\{0,y\}+\{0,z\}$ must be isomorphic to the direct sum $\{0,x\}\oplus\{0,y\}\oplus\{0,z\}$. And each one of those groups is isomorphic to $\mathbb{Z}_2$ (being of prime order $2$) and thus the final answer is $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus \mathbb{Z}_2$.
Note that you don't even need the fact that "$G$ is of order $8$ so it has to be one of the $3$ possible abelian groups of order $8$".

So yeah, except for my nitpicking your reasoning is correct. :)
