I am having trouble proving Fermat's little theorem using group theory. These are my steps so far:
Let $(Z_p,+,\times)$ a Field, and we assume we know that the non-zero elements form a Group multiplicatively of order $p-1$, and we know that the order for all $a$ in the group $(Z_p,\times)$, is $x$ such that $a^x\equiv 1\pmod p$. We need to show that $x=p-1$. Well, the subgroup generated by $a$ of order $x$ must divide the order of the group $p-1$, so $x|p-1$ by Lagrange, and so $p-1=x*m$ for some $m$ positive integer. So $a^{p-1}=a^{x*m}=1^m=1\pmod p$.
This proof feels sloppy and even wrong, but it's the best way I can word out my intuition, may you please help me out?
This is basically correct, but some of your wording is confusing and can be cleaned up. Specifically, the following part:
Here you are being very ambiguous about what $x$ represents or what you want to be true of $x$. Is $x$ supposed to be the order of every element, or just of one particular element $a$? Also, if $x$ is the order of $a$, then $a^x\equiv 1\pmod p$, but the converse is not true. So the property that $a^x\equiv 1\pmod p$ does not uniquely define the order as you seem to be implying in the first sentence. On the other hand, if you are defining $x$ as the order, then the second sentence is just wrong: you don't want to prove that $x=p-1$. Instead, you want to show that $a^{p-1}\equiv 1\pmod p$, which does not necessarily mean that $x=p-1$.
I might rephrase this part as follows:
In particular, notice how my first sentence clearly and unambiguously introduces a specific element $a$ and defines $x$ as the order of $a$, rather than presenting a jumble of facts about $x$ indirectly. Only once the definition is made do I start discussing other facts.