Vivek Jain, a postdoc in Allahabad, sent me a question. Is the following true?
Let G be a finite group and H be a proper non-trivial subgroup of G such that the core of H in G is trivial. Then there exist a right (or left) transversal of H in G which will generate the group.
It turns out that it is true, and the proof uses one of my favourite theorems.
First, the core of a subgroup H of G is the largest normal subgroup of G contained in H, that is, the intersection of all the conjugates of H; we say that H is core-free if the core is the trivial group. If this is the case, then the action of G on the set of right cosets of H by right multiplication is faithful, and embeds G as a subgroup of the symmetric group Sn, where n is the index of H in G. So it is really a question about permutation groups, which of course pleases me!
A subset S of a group G is said to be independent if no element of S lies in the subgroup generated by all the others. A lovely theorem of Julius Whiston, in Journal of Algebra (2000), says that the size of an independent set in Sn is at most n–1 (and, in fact, equality implies that the set generates Sn).
Now suppose that G is a counterexample to Jain’s statement, so that no right transversal of H generates G. Choose a transversal T which generates a subgroup K of G of largest possible order. Then |T|=n. So, by Whiston’s Theorem, T is not independent; thus we may omit an element t and still generate K. Now replace t in the transversal by an element in the same right coset of H but not in the subgroup K; the new transversal generates a larger subgroup, contrary to assumption. The proof is done.
One thing I like about this is that it is a theorem in pure group theory which is proved in the “right” way using permutation group theory (and, to repay the compliment, Whiston uses a lot of group theory in the proof of his theorem).