No group has $\mathbf{N}_5$ as its lattice of subgroups Let $\mathbf{N}_5$ be the lattice in the picture:

I'm asked to show that there is no group having this lattice as its lattice of subgroups.
To start with, I know that infinite groups can be excluded, but this doesn't take me very far...
My first thought was that, if $B \leq A$, then $BC \cap A = B(C \cap A)$.
Since in this case $A \cap C = \{1\}$, this yields $BC \cap A = B$.
I'm not sure this is any useful, because $BC$ doesn't have to be a subgroup (if it was, it would follow that $BC=B$, whence $C \subseteq B$, a contradiction).
This also doesn't seem to follow by Lagrange's Theorem alone.
I think it must be that $A = \langle a \rangle$ and $B = \langle a^r \rangle$, because from $A = \langle a, b \rangle$ and $B=\langle b \rangle$, it should follow that $\langle a \rangle$ is yet another subgroup of $A$.
But then, taking $C = \langle c \rangle$, it's not clear why can't $\langle a^r,c \rangle = G$.
In a tentative of showing this, I used the previous paragraph to claim that for $x \in A \setminus B$, there doesn't exist $b \in B$ and $c \in C$ such that $x = bc$.
Now taking $x = a \in \langle a^r,c \rangle = B \vee C$,
$$a = a^{rs_1}c^{k_1}\cdot \cdots \cdot a^{rs_n}c^{k_n},$$
where $s_i$ and $k_i$ are any integers.
Once again, if $a$ and $c$ don't commute, this doesn't seem to lead to any contradiction.
 A: Let the group be $G$.
$B$ and $C$ cannot be conjugate as $B$ has a proper over-group and $C$ does not. So each is normal,  and so we have a subgroup $BC$ of order $|B||C|$ containing $B$ and $C$. From the  lattice $BC=G$.
$A$ and $C$ cannot be conjugate as $B$ has a non-trivial subgroup and $C$ does not. So $A$ is also normal,  and so we have a subgroup $AC$ of order $|A||C|$ containing $A$ and $C$. From the  lattice $AC=G$.
Hence $|G|=|A||C|=|B||C|$. But $|A|\not=|B|$ so we have a contradiction.
A: The group $A$ has a single non-trivial subgroup $B$, so it must be cyclic. More precisely, we can say that $|A|=p^2$ for some prime $p$ for otherwise it would have more non-trivial subgroups. Similarly, $C$ is cyclic and its order is a prime $q$.
If $C$ normalizes $B$, then $N_G(B)$ contains both $A$ and $C$, so $B\unlhd G$.
In this case $BC$ is a subgroup, and your argument kicks in.
If $C$ does not normalize $B$, then $B$ has at least $q$ conjugate subgroups in $G$. They are all minimal (of order $p$) and all of them are contained in the corresponding conjugate of $A$, but there are no candidates for those conjugates of $A$ other than $A$ itself. So $A$ would need to contain $q$ non-trivial subgroups. A contradiction.

I'm fairly sure that a simpler argument exists :-) 
A: Assume $G$ is a finite group that has this as its subgroup lattice.
$B$ and $C$ are minimal subgroups so they're $\langle a\rangle, \langle b\rangle$ for some $a,b$.
Moreover $C$ has no nontrivial subgroup so $b$ has a prime order, similarly for $a$.
$A$ has only one subgroup hence it must be cyclic of order $p^2$ for some prime $p$; thus this prime $p$ is the order of $a$.
If the orders of $a$ and $b$ are the same prime $p$, then $G$ is a $p$-group, hence it has nontrivial center $Z(G) = A,B,C$ or $G$. If the center is $C$, $G/Z(G)=G/C$ is isomorphic to $A$ so it is cyclic so $G$ is abelian: a contradiction. Similarly if the center is $A$. So the center can only be $B$ or $G$.  If it's $G$, then $G$ is an abelian $p$-group of exponent $p^2$ : it's a product of $\mathbb{Z/pZ}$'s and at least one $\mathbb{Z/p^2Z}$. There can't be no $\mathbb{Z/pZ}$ so there is an antichain of size at least $3$ ($0\times \mathbb{Z/p^2Z}, \mathbb{Z/pZ}\times 0, \{(x,px), x\in [0,p-1]\}$) which is absurd. 
Hence the center is $B$. But then $BC\simeq B\times C$ is $G$  which implies that $G$ is abelian, a contradiction.
Hence the orders of $a,b$ are different primes, say $p,q$. But then $A$ is the only $p$-Sylow subgroup of $G$ and $C$ is the only $q$-Sylow subgroup of $G$ and both are normal and cyclic, hence $G$ is their product. But then $B\times C$ is a subgroup that's not in the lattice : a contradiction.
We've found a contradiction in every situation : there cab't be such a group
A: $B$, $C$ have no non-trivial proper subgroups, hence must by cyclic of prime order $p,q$, respectively.
$A$ must also be cyclic (or it would have two non-linked subgroups) and of order $p^2$ (or its order would have additional divisors leading to additional subgroups). 
From Sylow's theorems, we conclude $p\ne q$. Let $g$ be any of the $p^2q-p^2-q+1=(p^2-1)(q-1)>0$ elements of $G\setminus (A\cup C)$.
It follows that $\langle g\rangle=G$, so $G$ is cyclic of order $p^2q$ and has an additional uncharted subgroup of order $pq$.
