Let be the group of order 16 with the following presentation: . M is sometimes called the modular group of order 16, and has three subgroups of order 8: , , and . Every proper subgroup is contained in one of these three.
 Prove that , , and .
 Show that the lattice of subgroups of is isomorphic to the lattice of subgroups of (see the previous exercise) but that these two groups are not isomorphic.
Lemma 0: mod if and only if mod . Proof: We have mod if and only if for some , if and only if , if and only if mod .
Lemma 1: Let be a group and . If and , then and . Proof: We have , and . So . Similarly .

Claim 1: Let and . Then given by is an injective homomorphism. Proof: (Well defined) Suppose ; then mod 2 and mod 4, and by the lemma mod 8. Thus and , and . (Homomorphism) Let . Then . (Injective) Suppose . Then , so that . By Lemma 1, , so that mod 2, and , so that using Lemma 0, mod 4. Thus .
Claim 2: Let be the mapping defined in Claim 1.1. Then . Proof: Note that since and . So . Now let ; then for some and , so .
Since is an element of order 8 in , we have .
Claim 3: has order 8. Proof: The powers of are , , , , , , , and .
Since is an element of order 8 in , we have .
 We already know the subgroup lattices of , , and , the three maximal subgroups of . Moreover, every subgroup of 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 is as follows.
We can easily see that this lattice is isomorphic to the subgroup lattice of we computed in a previous exercise.
However, these two groups are clearly not isomorphic; is abelian, while is not.
Comments
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)
Thanks!
If we are allowed to use the Fundamental theorem of finitely generated abelian groups, it will be much easier to check that , since $uv^2=v^2 u$ and the largest order of an element in is 4.
typo: $uv^2=v^2 u$ to be