## 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.

• 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)

• nbloomf  On October 10, 2011 at 8:17 am

Thanks!

• 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 $uv^2=v^2 u$