Why isn't $\mathbb{Z}_2\oplus\mathbb{Z}_2=\{0,1,0+1\}=\{0,1\}=\mathbb{Z}_2\;?$ Let $x,y$ be two different transformations of order $2$ (i.e., they are reflections). Then we can form the two (abelian) groups 
$$(X,+)=\{0,x\} \quad \quad \text{and}\quad\quad(Y,+)=\{0,y\}$$
both of which are isomorphic to $\mathbb{Z}_2.$ 
Now define 
$$(G,+)=\{0,x\}\oplus \{0,y\}=\{0,x,y,x+y\},$$
where $\oplus$ is the direct sum. 
As far as I understand, $G$ is isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2.$
But wouldn't $$\mathbb{Z}_2\oplus\mathbb{Z}_2=\{0,1,0+1\}=\{0,1\}=\mathbb{Z}_2\;?$$
I guess the problem is that I can tell the difference between $x$ and $y$ in the case of $\{0,x\}\oplus \{0,y\}$ while I cannot in the case of $\mathbb{Z}_2\oplus\mathbb{Z}_2.$ 
This is probably elementary, but I'd really appreciate if someone would explain where I'm mistaken. Thanks. 
 A: The assertion that
$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 = \{0,1,0+1\} = \{0,1\}$$
is incorrect.
The idea of the direct sum of two groups is that we formally distinguish the elements of the first summand from those of the second, and apply the group operation to each component separately.
Formally, the (external) direct sum of two groups $(G,+_G)$ and $(H,+_H)$ is defined as the collection of pairs
$$ G \oplus H = \{(g,h) \mid g \in G, h \in H\}$$
with group operation defined by
$$ (g_1, h_1) +_{G\oplus H} (g_2, h_2) = (g_1 +_G g_2, h_1 +_H h2).$$
A: We have:
\begin{align}
\mathbb Z_2\oplus\mathbb Z_2
&= \mathbb Z_2^{1\rm st}\oplus \mathbb Z_2^{2\rm nd}\\
&= \{0^{1\rm st},1^{1\rm st}\}\oplus\{0^{2\rm nd},1^{2\rm nd}\}\\
&= \{0^{1\rm st}+0^{2\rm nd}, 0^{1\rm st}+1^{2\rm nd}, 1^{1\rm st}+0^{2\rm nd}, 1^{1\rm st}+1^{2\rm nd}\}
\end{align}
Note that at this stage, the "$+$" notation in those elements is a mere convention, that will be justified afterwards. That is, we actually define the elements to be the complete expressions $x^{1\rm st}+y^{2\rm nd}$. Also note that addition is by definition component-wise, that is,
$$(a^{1\rm st}+b^{2\rm nd}) + (c^{1\rm st}+d^{2\rm nd})
= (a+c)^{1\rm st}+(b+d)^{2\rm nd}$$
Now we see that $0^{1\rm st}+0^{2\rm nd}$ is the neutral element, so we just denote it as $0$. Moreover we notice that $\{0^{1\rm st}+x^{2\rm nd}:x\in \mathbb Z_2^{2\rm nd}\}$ is isomorphic to $\mathbb Z_2^{2\rm nd}$, therefore we omit the $0^{1\rm st}$, and analogously for $0^{2\rm nd}$.
Therefore we arrive at
$$\mathbb Z_2\oplus\mathbb Z_2 = \{0, 1^{1\rm st},1^{2\rm nd}, 1^{1\rm st}+1^{2\rm nd}\}$$
Note however that $1^{1\rm st}\ne 1^{2\rm nd}$, therefore you cannot simplify this to $\{0,1,1,1+1\}=\{0,1\}$
Also note that now the "$+$" notation for the elements is justified after the fact: $x^{1\rm st}$, when standing alone, is just a shorthand for the expression $x^{1\rm st} + 0^{2\rm nd}$, and $y^{2\rm nd}$ is just a shorthand for the expression $0^{1\rm st}+y^{2\rm nd}$. But if we interpret $x^{1\rm st}+y^{2\rm nd}$ not as a single expression, but as the sum of the standalone values $x^{1\rm st} + 0^{2\rm nd}$ and $0^{1\rm st}+y^{2\rm nd}$, then we get
$$(x^{1\rm st} + 0^{2\rm nd}) + (0^{1\rm st}+y^{2\rm nd}) =
(x+0)^{1\rm st}+(0+y)^{2\rm nd} = x^{1\rm st}+y^{2\rm nd}$$
so we recover the original expression, which justifies the notation.
A: Conside $G = ${$e, a, b, ab | a^2=e, b^2=e$}
Then $G$ is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$
Let $f: G \to \mathbb{Z}_2 \oplus \mathbb{Z}_2$ by
$f(e) = (0,0) $
$f(a) = (1,0) $
$f(b) = (0,1) $
$f(ab) = (1,1) $
$f(ab) = (1,1) = (1,0) + (0,1) = f(a) + f(b)$
So $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ cannot be equal to $\mathbb{Z}_2$.  
I think the issue is that your definition of Direct Sum is missing something important.
