Let be an odd prime, let , and let be the nonabelian group of order which has an element of order . Prove that and have isomorphic subgroup lattices.
The nontrivial subgroups of have order or . By §6.1 #28, we know that has precisely subgroups of order . By this previous exercise, also has precisely subgroups of order .
Note that is maximal in and elementary abelian. We claim that this is the unique elementary abelian maximal subgroup of . To see this, note that contains elements of order . Including the identity, there are exactly elements in such that . The set of these elements is a subgroup, and moreover any other elementary abelian subgroup of rank 2 must coincide with this one. In particular, one maximal subgroup is isomorphic to and the rest must be cyclic, by the classification of groups of order . Label the elementary abelian maximal subgroup and the cyclic maximal subgroups . Each of the contains a unique subgroup of order , while contains such subgroups- that is, contains all of the order subgroups of . Moreover, we see that the order subgroup of each is ; that is, for all distinct and . Label the non-Frattini order subgroups for .
The subgroup lattice of is thus as follows.
- There is a unique subgroup of order ; namely .
- There are subgroups of order ; namely and , . Each of these is contained in .
- There are subgroups of order ; namely and , . Moreover, is contained in every maximal subgroup, while each is contained only in .
- There is a unique subgroup of order 1; namely 1. This subgroup is contained in each subgroup of order .
We claim that is the unique elementary abelian maximal subgroup of . To see this, we will find a presentation for and prove some basic facts using it.
From the discussion on page 183 in the text on order groups, we see that has the presentation . Clearly every element of this group can be written in the form , with and . There are such expressions, and since we know that has order , every element can be written in this form uniquely. We now prove a lemma.
Lemma: Let be a group. If such that and . Then for all nonnegative integers , we have . Proof: If , the result is clear. Suppose ; we proceed by induction on . for the base case, note that . Similarly, if we have , and if , then . For the inductive step, suppose the conclusion holds for some . Then . Since , and , all but the first two terms in the expansion of are zero, so that mod . (Note that we used a lemma from this previous exercise.) Thus this is equal to , and the conclusion holds for . Suppose now that ; we proceed by induction on . If , then . For the inductive step, suppose the conclusion holds for some . Then . Again, all but the first two terms in the expansion of are 0 mod , so that this element in fact equals . Thus the conclusion holds for all .
We can now show that contains exactly elements such that . If is arbitrary in , then by the lemma . (Here we use the fact that is odd.) Thus we must have . Thus if , mod and is unrestricted; there are such elements, and clearly every such element has order (or 1). Thus has exactly elements of order or 1, which necessarily comprise . Moreover, we see that any other elementary abelian subgroup of order is in fact equal to . Thus the remaining maximal subgroups are cyclic; label these through .
Now is an order subgroup, and is the intersection of all pairs of maximal subgroups. is in fact the unique order subgroup in each of the ; any other order subgroups are thus contained in (any two disjoint order subgroups have an elementary abelian product). has such subgroups; thus has order subgroups.
Thus we see that and have isomorphic subgroup lattices.