Inverses and orders in the group $Z_5$ Ill like some guidance to solve this kind of question ( I have many like this) and I have no clue what I need to do here:
I need to find 
$$g^{-1}$$
 and 
$$o(g)
$$
when $G$ is $Z_{_5}$.
adding - is I have G is $Z^x_{_5}$ what group is this????
 A: For each element $g \in G$, compute $g$, $g^2$, $g^3$, ... until you reach 1. The order is the number of steps you make; the inverse is the last element before you reach 1.
A: So $\mathbb{Z}_5 = \{[0], [1], [2], [3], [4]\}$ is an additive group (so we can/should write $-g$ for $g^{-1}$). You should try to write out the whole addition table. That way you will be able to see what the inverse of an element is.
So for example, for $g = [2]$ you have that $[2] + [3] = [5] = [0]$, hence the inverse of $[2]$ is $[3]$.
To find the order of an element you keep adding it to itself until you reach $[0]$. So for example of $g = [3]$, then you have $[3] + [3] + [3] +[3] + [3] = [15] = [0]$. So the order of $[3]$ is $5$. (Note here that adding $[3]$ to itself less times will not give you $[0]$.
Added: You ask about the group $\mathbb{Z}^{\times}_5$. It is a multiplicative group. We take the additive group and remove $[1]$. So we have
$$
\mathbb{Z}_5^{\times} = \{[1], [2], [3], [4]\}.
$$
Now the composition of elements is multiplication. So we have for example that $[3]\cdot[4] = [12] = [2].$ Or we have that  $[2]^{-1} = [3]$, because $[2]\cdot [3] = [2\cdot 3] = [6] = [1]$. For the order of an element we now just use multiplication. So for example if $g = [2]$, then $[2]\cdot[2]\cdot[2] = [6] = [1]$, so the order of $[2]$ is 3 (again noting that any power  of $[2]$ less than $3$ will not give $[1]$). 
Another note is that taking the units (i.e. putting the ${}^\times$ on $\mathbb{Z}_n$) doesn't always mean that you just take away zero. You are simply asking for all the elements that have a multiplicative inverse. As suggested in the comments to your question, you can take a look at the Wikipedia article about the multiplicative group.
