Only one group having $\mathbf{M}_3$ as its lattice of subgroups The problem.
Consider the lattice $\mathbf{M}_3$ below

Clearly, this is the lattice of subgroups of $\mathbb{Z}_2^2$ (isomorphic to the Klein four-group $\mathbf{V}_4$), taking $A = \langle (0,1) \rangle$, $B = \langle (1,0) \rangle$ and $C = \langle (1,1) \rangle$.
I want to show that, up to isomorphism, this is the only group $G$ having $\mathbf{M}_3$ as its lattice of subgroups.
My solution.
To start with, clearly $G$ is finite.
In this case, since 
$$A \cap B = B \cap C = C \cap A = \{1\},$$
we have that
$$|AB| = |A|\cdot |B|, \quad |BC| = |B|\cdot|C|, \quad |CA|=|C|\cdot|A|,$$
and
$$|G| = (|A|-1) + (|B|-1) + (|C|-1) + 1 = |A|+|B|+|C|-2.$$
Since $A,B,C$ are minimal, they are cyclic of prime orders, $p_A, p_B, p_C$.
First case.
Suppose that $p_A, p_B, p_C$ are all different.
As conjugate subgroups are isomorphic, it follows that $A,B,C \triangleleft G$, whence $AB, BC, CA \leq G$ and thus, in our specific case, $AB=BC=CA = G$, yielding
$$|G| = p_Ap_B = p_Bp_C = p_Cp_A,$$
whence $p_A=p_B=p_C$, a contradiction with the hypothesis.
Second case.
Suppose $p_A \neq p_B=p_C$.
In this case $A \triangleleft G$, whence $AB, AC \leq G$ and $|G| = |A|\cdot|B| = p_Ap_B$.
By the First Theorem of Sylow, $G$ has a Sylow $p_B$-subgroup $P$ of order $p_B$.
Necessarily $P=B$ or $P=C$ and since $|B|=|C|$, they both are Sylow  $p_B$-subgroups of $G$ and $n_{p_B}=2$.
By the Third Theorem of Sylow, $n_{p_B} | p_A$, whence $p_A=2$ and $|G|=2p_B$. 
So 
$$p_B^2 = |BC| \leq 2p_B,$$
whence $p_B \leq 2$ and thus, $p_A=p_B=p_C=2$, again, a contradiction.
Third case.
Suppose that $p_A=p_B=p_C=p$.
If $q \neq p$ is another prime number dividing $|G|$, then $G$ has a Sylow $q$-subgroup, which is a contradiction because all proper non-trivial subgroups of $G$ have order $p$.
Hence
$$|G|=p^r,$$
for some $r>1$ (it can't be $r=1$ because $G$ has non-trivial proper subgroups).
$$r \geq 2 \quad\text{because}\quad p^2 = |AB| \leq |G|,$$
$$r \leq 3 \quad\text{because}\quad |G| \leq |A|\cdot |B|\cdot|C|.$$
Recall that, in this case, $|G|=3p-2$.
If $|G|=p^3$, then
$$0 = p^3 -3p + 2 = (p+2)(p-1)^2,$$
whence $p=-2$, which is, of course, impossible;
or $p=1$ which is not a prime (and $G$ would be trivial).
Therefore $|G|=p^2$ and 
$$0 = p^2 - 3p + 2 = (p - 1)(p - 2),$$
yielding $p=2$, since, again, $G$ is not trivial.
So $|A|=|B|=|C|=2$ and thus $G$ is an exponent $2$ group, whence abelian and having order $4$.
This is enough to fully determine $G$.
My questions.


*

*I suspect that there is a much simpler proof, without going through all those cases and using elementary group theory (but I'm not claiming this one's advanced...). I would welcome a concise (and elementary) proof even if you don't want to take the time to read through all the above.

*Anyway, is this all right?


Thanks in advance!
 A: I think what you say is right. Let me offer you a different approach which would let you attack more general cases when $3$ is replaced by something larger e.g 4, or ...or  8, or whatever you care to try.
As you know the only groups with no non-trivial proper subgroups are cyclic of prime order. So we may assume $|A|=p$, $|B|=q$, $|C|=r$, where $p,q,r$ are primes and without loss of generality $p\geqslant q \geqslant r$.
Now $A$ acts by conjugation on the set of minimal subgroups of our group $G$. It fixes $A$ of course, and permutes the others in orbits of length $1$ or $p$. 
Suppose that $p>2$. Then $A$ has to fix each of $B$ and $C$ since $2<p$. Now $A$ will act on the elements of $B$ by conjugation: it will fix the identity $1$ and must permute the $q-1$ others in orbits of length $1$ or $p$. As $q-1<p$ it will therefore fix each of them. That is, if we take generators of $A$ and $B$ we have elements of order $p$ and $q$ which commute; they generate a subgroup containing $A$ and $B$ which is therefore $G$. So $G$ is abelian of order $pq$.If $q=p$ then we have a contradiction: the cyclic group of order $p^2$ has only one non-trivial proper subgroup, and the elementary abelian group of order $p^2$ has $(p+1)$ subgroups of order $p$.
We therefore have that $p=2$ and so $p=q=r=2$. The order of $G$ is now $4$. The cyclic group has only one subgroup of order $p$, the Klein group has the structure we want.
