Order of Subgroup , and exponentiating elements 
I understand how to compute the orders but can someone explain why $ x $ = $ [1] $ goes to $ x^2 $ = $ [2] $ 
 A: The exponent, when the base is a group element, simply denotes the number of times  to perform the operation of the group on the given element. In the additive groups $\mathbb Z_n$, the group operation is addition modulo $n$. So $x^2 = x + x = 2x \pmod n\;$ e.g.  E.g., $[3]^2 = [3] + [3] \pmod 6 \equiv [0] \in \mathbb Z_6$. Indeed, we have that the order of $[3]$ in $\mathbb Z_6$ is equal to $2$.
This is true for additive groups in general. For example, in the group of integers with addition, $x^n$ would be $\;nx$: the sum of adding $x$ to itself $n$ times.
Whatever the operation that in part defines a group, let's just denote that operation $*,\;$ then if $g$ is an element of the group, we have $\;g^2 = g*g\;$. So for addition, that's $g + g$, for multiplication as we normally define it, $g^2 = g\cdot g$. 
For composition of, say, permutations in $S_n$, if $\alpha \in S_n$, then $\alpha^2 = \alpha\circ \alpha.$
And so on...It's very typical/standard, instead of writing $\underbrace{g * g * g *\cdots}_{\large n \text{times}}$, to write $g^n$.  By doing this, we can speak about groups in general, without needing to know the precise operation of a group.
A: The operation of the group is addition.  Then:
If $x = [1]$, then $x^{2} = x + x = [1] + [1] = [2]$.
Similarly, if $y = [2]$, then $y^{2} = y + y = [2] + [2] = [4]$.
