Exercise 2.5.12 (Lattice of subgroups of $Z_2\times Z_4$)

Answers

Proof. By definition, $A = Z_{2} \times Z_{4}$ , therefore $| A | = 8$ . We write $Z_{2} = ⟨ x ⟩ = {1, x}$ , and $Z_{4} = ⟨ y ⟩ = {1, y, y^{2}, y^{3}}$ , and we take $a = (x, 1)$ and $b = (1, y)$ the generators of $A$ which satisfy $a^{2} = b^{4} = (1, 1) = 1_{A}$ and $ab = ba = (x, y)$ . Let $1$ denote $1_{A} = ({[1]}_{2}, {[1]}_{4})$ .

First $⟨ a, b^{2} ⟩ = {1, a, b^{2}, a b^{2}}$ is a subgroup, and all elements in this subgroup have orders $1$ or $2$ : $a^{2} = 1$ , ${(b^{2})}^{2} = 1$ and ${(a b^{2})}^{2} = a^{2} b^{4} = 1$ . Therefore (see Ex. 10)

⟨ a, b^{2} ⟩ ≃ Z_{2} \times Z_{2} ≃ V_{4} .

Since $b^{4} = 1$ and $b^{2} = (y^{2}, 1) \neq 1$ , the order of $b$ is $4$ , therefore

⟨ b ⟩ ≃ Z_{4} .

Finally, ${(ab)}^{2} = a^{2} b^{2} = b^{2} = (1, y^{2}) \neq 1$ , and ${(ab)}^{4} = (1, y^{4}) = 1$ , therefore $| ab | = 4$ , so

⟨ ab ⟩ ≃ Z_{4} .

Let $S$ be a proper subgroup of $A$ . By Lagrange’s Theorem $| S | = 1$ or $| S | = 2$ or $| S | = 4$ .

If $| S | = 1$ , then $S = {1}$ is included in any subgroup.

Since

\begin{array}{l} A & = {1, b, b^{2}, b^{3}, a, ab, a b^{2}, a b^{3}}, \\ H = ⟨ a, b^{2} ⟩ & = {1, a, b^{2}, a b^{2}}, \\ K = ⟨ b ⟩ & = {1, b, b^{2}, b^{3}}, \\ L = ⟨ ab ⟩ & = {1, ab, b^{2}, a b^{3}}, \end{array}

then

A = ⟨ a, b ⟩ \cup ⟨ b ⟩ \cup ⟨ ab ⟩ .

Therefore every element of order $2$ is in one of these $3$ subgroups, thus every subgroup $S$ of order $2$ is included in one of these subgroups.

Finally, suppose that $| S | = 4$ . Then $S$ is cyclic or isomorphic to $V_{4}$ . If $S = ⟨ h ⟩$ , since $h \in H \cup K \cup L$ , then $S \leq H$ or $S \leq K$ or $S \leq L$ .

If $S ≃ Z_{2} \times Z_{2}$ , then $S = ⟨ h, k ⟩$ where $| h | = | k | = 2$ and $h \neq k$ . Since the only elements of order $2$ are $a, b^{2}, a b^{2}$ , the only possibilities are

{h, k} \in {{a, b^{2}}, {a, a b^{2}}, {b^{2}, a b^{2}}} .

But $H = ⟨ a, b^{2} ⟩ = ⟨ a, a b^{2} ⟩ = ⟨ b^{2}, a b^{2} ⟩$ (any pair of nontrivial elements of $H$ generates $H$ ), so $S = H$ .

Every proper subgroup is contained in $H$ , $K$ or $L$ , which are consequently the only maximal subgroups of $A$ .

This gives the lattice of subgroups of $A$ :

□

I’m trying to learn how to use "Gap". This gives the lattice of the subgroups of $Z_{2} \times Z_{4}$ :

gap> K8 := Group([(1, 2), (3, 4, 5, 6)]);;  Order(K8);
8
gap> couple := GeneratorsOfGroup(K8);;
gap>  a := couple[1];; b := couple[2];;
gap>  Print("a = ",a, ", b = ", b);
a = (1,2), b = (3,4,5,6)
gap> for i in [0..1] do
>        for j in [0..3] do
>            Print("a^",i,"b^",j," = ", a^i*b^j,"\n") ;
>        od;
>     od;
a^0b^0 = ()
a^0b^1 = (3,4,5,6)
a^0b^2 = (3,5)(4,6)
a^0b^3 = (3,6,5,4)
a^1b^0 = (1,2)
a^1b^1 = (1,2)(3,4,5,6)
a^1b^2 = (1,2)(3,5)(4,6)
a^1b^3 = (1,2)(3,6,5,4)
gap> T:=Group(());;
gap> L:=IntermediateSubgroups(K8,T);;
gap> for G in L.subgroups do
>     Print(Order(G)," : ",G,"\n");
> od;
2 : Group( [ (1,2) ] )
2 : Group( [ (3,5)(4,6) ] )
2 : Group( [ (1,2)(3,5)(4,6) ] )
4 : Group( [ (3,5)(4,6), (1,2) ] )
4 : Group( [ (3,4,5,6), (3,5)(4,6) ] )
4 : Group( [ (1,2)(3,4,5,6), (3,5)(4,6) ] )
gap> Print(L.inclusions);
[ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 2, 4 ], [ 3, 4 ], [ 2, 5 ], [ 2, 6 ],
  [ 4, 7 ], [ 5, 7 ], [ 6, 7 ] ]

This confirms the previous pattern.

richardganaye

2025-11-06 09:30

Exercise 2.5.12 (Lattice of subgroups of $Z_2\times Z_4$)

Answers

Comments

Add answer