Showing that Ab(G) need not be complete

Question

What's the easiest example to show that $Ab(G)$, the set of Abelian subgroups of a group $G$, need not be complete? I heard that $D_4$ was a good example, but $Ab(D_4)=Sub(D_4)$, which is complete.

Edit: I am referring to the partial ordering of $Ab(G)$ and $Sub(G)$ by inclusion. When I say a partially-ordered set $P$ is complete, I mean that every subset $C \subset P$ has a least upper bound, under the ordering.

Edit: When I say $D_4$, I am referring to the dihedral group of order $4$, which is isomorphic to $C_2 \times C_2$, rather than the group of symmetries of a square. This usage is standard in my institution. Therefore $D_4$ is abelian.

Answer 1

$Ab(D_4)$ (where $D_4$ is the symmetry group of the square; perhaps you call this $D_8$) is not complete. If we present $D_4$ as $\langle a,b \mid a^4=b^2=e, ab=ba^{-1} \rangle$, then $\langle a \rangle$ and $\langle b \rangle$ are two abelian subgroups without a common upper bound in $Ab(D_4)$.

Answer 2

In the nonabelian group of order $p^3$ and exponent $p$ ($p$ an odd prime), which is isomorphic to the multiplicative group of unimodular ($1$'s in the diagonal) $3\times 3$ matrices with coefficients in $\mathbb{F}_p$, every maximal subgroup has order $p^2$ and is abelian. But it has more than one maximal subgroup: for example, the subgroup of all matrices of the form $$\left(\begin{array}{ccc} 1 & 0 & *\\ 0 & 1 & *\\ 0 & 0 & 1 \end{array}\right)$$ is maximal and abelian, as is the subgroup of matrices of the form $$\left(\begin{array}{ccc} 1 & * & *\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right).$$ But the two are not comparable, and there is no abelian subgroup that contains both.

By the way: a presentation for this group is given by $$\Bigl\langle x,y,z\Bigm| x^p = y^p = z^p = 1,\ xz=zx,\ yz=zy,\ yx=xyz\Bigr\rangle$$ with $x$ corresponding to the matrix with a $1$ in the $(1,2)$ entry; $y$ to the matrix with a $1$ in the $(2,3)$ entry; $z$ to the matrix with a $1$ in the $(1,3)$ entry. If you replace $p$ with $2$, you get $$\Bigl\langle x,y,z\Bigm| x^2 = y^2 = z^2 = 1, xz=zx,\ yz=zy,\ yx=xyz\Bigr\rangle$$ which is isomorphic to the dihedral group with $8$ elements, $D=\langle r,s\mid r^4 = s^2 = 1,\ sr=r^3s\rangle$, by the map that sends $z$ to $r^2$, $x$ to $s$, and $y$ to $sr$.