On $2$-groups with a property If $G$ is a non-abelian $p$-group ($p>2$) such that any two maximal cyclic subgroups have trivial intersection, then $G$ is of exponent $p$ (see "Groups of Prime Power Order-1"- Berkovich, Exer. 2, p.27). Here, "maximal cyclic subgroups" means "maximal among cyclic subgroups".
For $p=2$, the situation is different. The dihedral $2$-groups have property that any two maximal cyclic subgroups have trivial intersection, but exponent is larger. The natural question is, are these only $2$-groups with such property?
 A: Let $X$ be the class of finite $2$-groups such that every pair of maximal cyclic subgroups intersect trivially.
Proposition: $X$ consists of the elementary abelian, the cyclic, and the dihedral $2$-groups.
Proof: The abelian groups in $X$ are the cyclic $2$-groups and the elementary abelian $2$-groups.
$X$ is closed under subgroups, so if we take $G$ from $X$ of minimal order amongst the non-abelian non-dihedral groups, then every subgroup of $G$ is cyclic, elementary abelian, or dihedral.
If $G$ has a maximal cyclic subgroup then direct inspection of the classification of such (Theorem 1.2) yields $G$ is dihedral, a contradiction.
Hence every maximal subgroup of $G$ is dihedral or elementary abelian.
By theorem 12.12.c the number of maximal subgroups that are dihedral is either 0 or 4, and by theorem 9.10, the number of maximal subgroups is 7, hence $G$ has a maximal subgroup $N$ that is elementary abelian.
Hence $G=\langle x,N\rangle$ for $x$ of order 4. $1<Z(G) \leq C_G(N) = N$ so let $1 \neq z \in Z(G) \leq N$. Then $(xz)^2 = x^2 z^2 = x^2 \in \langle x \rangle \cap \langle xz \rangle$ so $x^3 = xz$ and $z=x^2$ is the unique non-identity element of $Z(G)$.
Note that $[G,G] \leq N$ is abelian.
Consider $(xy)^2$ for $y \in N$: $(xy)^2 = xyxy = xxy[y,x]y = xxyy[y,x] = z[y,x]$ wher the second to last equality follows because $y, [x,y] \in N$ and the last because $y \in N$ (so $y^2=1$) and $x^2=z$.
Now consider $(xyz)^2 = z [yz,x] = z[y,x] \in \langle xy \rangle \cap \langle xyz \rangle$. Hence either $(xy)^3 = xyz$ so that $(xy)^2 = z$ so that $[y,x]=1$ or $z[y,x]=1$ so $[y,x]=z$.
We draw two conclusions: (1) $(xy)^2$ is either equal to $zz=1$ or $z1=z$, so that $G^2 = \langle z \rangle$, and (2) $[x,y] = 1$ or $[x,y]=z$ so $[G,G] = \langle z \rangle$.
In particular, $\Phi(G) = [G,G] = Z(G) = \langle z \rangle$ so $G$ is extraspecial. However, in an extraspecial group of order greater than $p^3$, $z$ will be contained in multiple cyclic subgroups, a contradiction. $\square$
