Morley rank of group

For the first question: $M$ is an abelian group of exponent $4$, and hence so is $M^*$. It follows that $2M^*$ is an abelian group of exponent $2$. An abelian group of prime exponent $p$ is a vector space over the $p$-element field. (More generally, an abelian group of exponent $n>0$ is a $mathbf Z/nmathbf Z$-module.)

Regarding your second question, I suppose you want to compute the Morley rank of $M$ as a pure abelian group. The subgroup $2Mleq M$ is infinite, so it has Morley rank at least $1$, and $[M:2M]$ is infinite, so $M$ has Morley rank at least $2$.

To obtain the upper bound, it is enough to show that $M/2M$ and $2M$ are both rank $1$. But that follows from the fact that (for each of them), the induced structure is that of a vector space over the two-element field, and as such, they are strongly minimal.

There are two ways to interpret your first question. One is purely algebraic and straightforward: the group $2M^{*}$ has exponent $2$ (because for all $ain 2M^*$, $a = 2b$ for some $bin M^*$, so $2a = 4b = 0$), and every group of exponent $2$ admits a unique structure as a vector space over $mathbb{Z}_2$.

The second way to interpret your question is model-theoretic and a bit more complicated: the induced structure on $2M^{*}$ from the whole model $M^*$ is the same as (interdefinable with) the vector space structure on $2M^*$. Clearly the vector space operations on $2M^*$ are definable in $M^*$. In the other direction, any subset of $2M^*$ which is definable in the abelian group language with parameters from $M^*$ is also definable in the vector space language with parameters from $2M^*$. This follows abstractly from the fact that $M^*$ is stable (like every abelian group), so $2M^*$ is stably embedded in $M^*$, or you can argue directly using quantifier elimination.

For your second question, let's compute the Morley rank of $M^*$. It's easy to see that the Morley rank is $geq 2$: The subgroup $2M^*$ has infinitely many disjoint cosets. Each of these cosets is definable and has Morley rank $geq 1$ (since it is inifnite), so they witness $text{MR}(M^*)geq 2$.

For the converse, we need to understand all definable sets in $M^*$, so you should start by showing that the theory has quantifier elimination (it may be useful to start with the usual quantifier elimination down to positive primitive formulas in theories of abelian groups, and then eliminate the last existential quantifier block).

An atomic formula in one free variable is equivalent to $nx + a = 0$, where $n in {0,1,2,3}$ and $ain M^*$. If $n = 0$, this formula defines $M^*$ if $a = 0$ or $varnothing$ if $aneq 0$. If $n = 1$ or $3$, this formula defines a single element of $M^*$. Finally, if $n = 2$, this formula defines a coset of $2M^*$.

It follows from quantifier elimination that every definable subset of $M^*$ is either $M^*$ minus finitely many points or is a finite union of cosets of $2M^*$, plus or minus finitely many points. Since $2M^*$ is strongly minimal (its induced structure is a vector space, as shown above), it has Morley rank $1$, so every coset of $2M^*$ has Morley rank $1$, and hence every finite union of cosets (plus or minus finitely many points) also has Morley rank $1$. It follows that $M^*$ has Morley rank $leq 2$, since we cannot find an infinite family of (not even $2$!) disjoint definable sets of Morley rank $geq 2$.

The above argument works directly from the definition of Morley rank. In the other answer, tomasz gave a quicker solution using a fact about additivity of Morley rank in stable groups.