Compute the subgroup lattice of the modular group of order 16

Let M be the group of order 16 with the following presentation: M = \langle u,v \ |\ u^2 = v^8 = 1, vu = uv^5 \rangle. M is sometimes called the modular group of order 16, and has three subgroups of order 8: \langle u,v^2 \rangle, \langle v \rangle, and \langle uv \rangle. Every proper subgroup is contained in one of these three.

  1. Prove that \langle u,v^2 \rangle \cong Z_2 \times Z_4, \langle v \rangle \cong Z_8, and \langle uv \rangle \cong Z_8.
  2. Show that the lattice of subgroups of M is isomorphic to the lattice of subgroups of Z_2 \times Z_8 (see the previous exercise) but that these two groups are not isomorphic.

Lemma 0: a = b mod n if and only if ka = kb mod kn. Proof: We have a = b mod n if and only if a-b = qn for some q, if and only if ka-kb = kqn, if and only if ka = kb mod kn. \square

Lemma 1: Let G be a group and x,y \in G. If \langle x \rangle \cap \langle y \rangle = 1 and x^ay^b = 1, then x^a = 1 and y^b = 1. Proof: We have y^b = x^{-a} \in \langle x \rangle, and y^b \in \langle y \rangle. So y^b = 1. Similarly x^a = 1. \square

  1. Claim 1: Let Z_2 = \langle x \rangle and Z_4 = \langle y \rangle. Then \varphi : Z_2 \times Z_4 \rightarrow M given by (x^a,y^b) \mapsto u^av^{2b} is an injective homomorphism. Proof: (Well defined) Suppose (x^{a_1},y^{b_1}) = (x^{a_2},y^{b_2}); then a_1 = a_2 mod 2 and b_1 = b_2 mod 4, and by the lemma 2b_1 = 2b_2 mod 8. Thus u^{a_1} = u^{a_2} and v^{2b_1} = v^{2b_2}, and \varphi(x^{a_1},y^{b_1}) = \varphi(x^{a_2},y^{b_2}). (Homomorphism) Let (x^{a_1},y^{b_1}), (x^{a_2},y^{b_2}) \in Z_2 \times Z_4. Then \varphi((x^{a_1},y^{b_1}) (x^{a_2},y^{b_2})) = \varphi(x^{a_1+a_2}, y^{b_1+b_2}) = u^{a_1+a_2} v^{2(b_1+b_2)} = u^{a_1}v^{2b_2}u^{a_2}v^{2b_2} = \varphi(x^{a_1},y^{b_1}) \varphi(x^{a_2},y^{b_2}). (Injective) Suppose \varphi(x^{a_1},y^{b_1}) = \varphi(x^{a_2},y^{b_2}). Then u^{a_1}v^{2b_1} = u^{a_2}v^{2b_2}, so that 1 = u^{a_2-a_1}v^{2b_2 - 2b_1}. By Lemma 1, u^{a_2-a_1} = 1, so that a_2 = a_1 mod 2, and v^{2b_2 - 2b_1} = 1, so that using Lemma 0, b_2 = b_1 mod 4. Thus (x^{a_1},y^{b_1}) = (x^{a_2},y^{b_2}). \square

    Claim 2: Let \varphi be the mapping defined in Claim 1.1. Then \mathsf{im}\ \varphi = \langle u,v^2 \rangle. Proof: Note that u, v^2 \in \mathsf{im}\ \varphi since u = \varphi(x,1) and v^2 = \varphi(1,y). So \langle u,v^2 \rangle \leq \mathsf{im}\ \varphi. Now let w \in \mathsf{im}\ \varphi; then w = u^av^{2b} = u^a(v^2)^b for some a and b, so w \in \langle u,v^2 \rangle. \square

    Since v is an element of order 8 in M, we have \langle v \rangle \cong Z_8.

    Claim 3: uv \in M has order 8. Proof: The powers of uv are uv, v^6, uv^7, v^4, uv^5, v^2, uv^3, and 1. \square

    Since uv is an element of order 8 in M, we have \langle uv \rangle \cong Z_8.

  2. We already know the subgroup lattices of \langle u,v^2 \rangle, \langle v \rangle, and \langle uv \rangle, the three maximal subgroups of M. Moreover, every subgroup of M appears in the lattice of one of these subgroups. By identifying subgroups which appear in more than one sublattice, we see that the subgroup lattice of M is as follows.

    We can easily see that this lattice is isomorphic to the subgroup lattice of Z_2 \times Z_8 we computed in a previous exercise.

    However, these two groups are clearly not isomorphic; Z_2 \times Z_8 is abelian, while M is not.

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  • Jeef  On October 9, 2011 at 7:48 pm

    Nice one.

    Typo alert: parenthesis before equal sign should be dropped on line nine of claim 1.

    u^{a_1}v^{2b_1}) = u^{a_2}v^{2b_2}

    (this one)

  • Gobi Ree  On November 17, 2011 at 8:13 pm

    If we are allowed to use the Fundamental theorem of finitely generated abelian groups, it will be much easier to check that \langle u,v^2 \rangle \cong  Z_2 \times Z_4, since $uv^2=v^2 u$ and the largest order of an element in \langle u,v^2 \rangle is 4.

  • Gobi Ree  On November 17, 2011 at 8:14 pm

    typo: $uv^2=v^2 u$ to be uv^2=v^2 u

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