Exercise 2.5.13 (Lattice of subgroups of $Z_2 \times Z_8$)

Simply reattach these three trees to obtain the lattice of subgroups of $G$.

Now we prove that the assertions of the statement are true: $G$ has order $16$ and has three subgroups of order $8$: $H=⟨x,y^2⟩\simeq Z_2\times Z_4,K=⟨y⟩\simeq Z_8$ and $L=⟨\mathit{xy}⟩\simeq Z_8$ and every proper subgroup is contained in one of these three.

First $G=Z_2\times Z_8$ has order $16$. Since $|y|=|\mathit{xy}|=8$, $K=⟨y⟩\simeq Z_8$ and $L=⟨\mathit{xy}⟩\simeq Z_8$. Moreover $⟨x⟩\cap ⟨y^2⟩=\{1\}$, where $|x|=2,|y^2|=4$ and $x,y^2$ commute, therefore $H=⟨x,y^2⟩\simeq Z_2\times Z_4$.

Finally, we show that every proper subgroup is contained in one of these three.

There are only three elements of order $2$, which are $x,x{y}^4$ and $y^4$. They belong to $⟨x,y^4⟩\le H$, so every subgroup of $G$ of order $1$ or $2$ is contained in $H$.

Now take a proper subgroup $S$ of order $|S|>2$, so that $|S|=4$ or $|S|=8$. We first show that $y^4\in S$. If not, since $y^4={(y^2)}^2={(y^3)}^4={(y^5)}^4={(y^6)}^2={(y^7)}^4$, then $y,y^2,y^3,y^4,y^5,y^6,y^7$ don’t belong to $S$, therefore, if we put $T=⟨y⟩$, then

Then $|\mathit{ST}|=|S||T|=8|S|>16$. This is impossible, because $\mathit{ST}\subseteq G$, where $|G|=16$. This contradiction shows that $y^4\in S$, so

Since $G$ is abelian, every subgroup is normal in $G$. Then $G∕⟨y^4⟩\simeq Z_2\times Z_4$. By Theorem 20 (the Fourth or Lattice Isomorphism Theorem), there are as many proper subgroups in $G$ which contain $⟨y^4⟩$ than proper subgroups of $Z_2\times Z_4$, that is $7$ subgroups. Since we know exactly $7$ subgroups that contain $⟨y^4⟩$ (i.e., $⟨y^4⟩$, $⟨x,y^4⟩$, $⟨x{y}^2⟩$, $⟨y^2⟩$, $⟨x,y^2⟩$, $⟨y⟩$, $⟨\mathit{xy}⟩$), there is no other proper subgroup of $G$ than these, and all are contained in $H$, $K$ or $L$.

In conclusion, every proper subgroup of $G$ is contained in $H$, $K$ or $L$, and the preceding lattice is complete.