Inverse of zero missing for all finite fields I am having a little touble with finite fields at the moment.
I am just working from a high school text wich says that the inverse of an element 
in a group is unique, which to me implies that all elements have an inverse.
But when I look a any multiplicative group $F_2$, the zero $0$ element never has an inverse?
I was wondering if I have something wrong or is zero special? Or is the theorem wrong? The thing that worries me is that if I treated groups abstractly, with symbols rahter than numbers, then 0 would look very strange, i.e. what would zero correlate isomorphically with symetries?
I think I should know this.
 A: You say that: 

the inverse of an element being unique implies that every element has an inverse. 

This is not true. For example, when you say: "Every married man has exactly one wife." it does not imply that every man is married. Likewise, what you wrote simply says that if an element has an inverse, then that inverse is unique. 
Then you ask:

is zero special? 

Yes, it is pretty special. It is the only element in a ring, for which $a+0=0+ a=a$, for all elements $a$ of the ring. Furthermore, every ring has a zero. 
I find it very difficult to understand what you are trying to say in your last sentence. You need to learn clearly the difference between groups, rings, and fields. Also, groups are  always considered "abstractly" and rarely ever with numbers (rather numerals). Groups will always have a neutral element. As a consequence of that, rings will always have a zero. You don't have to call it zero,  but there will always be an element $x$ in the ring such that, $a*x=x*a=a$ for all elements $a$ of the ring, where $*$ is one of the opererations defined on the ring. You may call it whatever you like. 

To see why $0$ never has a multiplicative inverse in a nontrivial ring, see my answer here.
A: The additive group of the field $F_q$ contains the $q$ elements $0,1,2,\ldots,q-1$. The multiplicative group of $F_q$ contains the $q-1$ elements $1,2,\ldots,q-1$ (and does not contain $0$).
In your example, the multiplicative group of $F_2$ just contains the single element $1$.
