Use the subgroup lattice of to find the centralizer of every element of .
Recall that . We computed the subgroup lattice of in a previous exercise.
Note that we can use the relations to write any word in in using only positive powers. Then we can use the relation to express every element of in the form for some and , and the first two relations give us that and . Since we know that the order of this group is 16, all such elements are distinct.
Lemma 1: Let be a group, , and . If for all , then . Proof: Every element of is a word in ; by hypothesis, then, commutes with every element of .
Lemma 2: . Proof: We have and . By Lemma 1, then, .
We will make judicious use of the lemmas of a previous exercise concerning centralizers.
We have , so is either or . But so . | ||
, so that is either , , , , or . Now and . | ||
Lemma 4 to a previous exercise. | ||
Lemma 1 to a previous exercise | ||
Lemma 1 to a previous exercise | ||
Lemma 4 to a previous exercise | ||
We have , but . | ||
, and . | ||
Lemmas 2 and 3 to a previous exercise. | ||
Lemma 1 to a previous exercise | ||
Lemma 4 to a previous exercise | ||
, but | ||
Lemma 4 to a previous exercise |
Comments
typo: ]tau — \tau,
— in the tau’s row of the table.
Thanks!