TITLE: No group has $\mathbf{N}_5$ as its lattice of subgroups
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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.