= Solution
\b[No field has six elements.] A <finite field> has positive <characteristic>: the sequence $0,1,2\cdot1,\ldots$ must repeat. By part (a), this characteristic is a <prime number> $p$. Translation by $1$ partitions the field into cycles of length $p$, since returning to a starting point after $k$ steps is equivalent to $k1=0$. Thus $p$ divides six and $p=2$ or $p=3$.
The multiples of $1$ form a copy of the <prime field> $\mathbb F_p$. For completeness, the inverse of a nonzero residue $a$ lies among these multiples: the integer <Bezout identity> $ab+cp=1$ gives $(a1)(b1)=1$. The entire field is a <vector space> over this subfield. A maximal <linearly independent set> is finite and spans, so it is a <basis> of some positive size $d$. Unique basis coordinates give exactly $p^d$ elements. This proves <finite field cardinality is a prime power> here. But six is neither a power of two nor a power of three, a contradiction.
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