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 *S*_{n}, 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 *S*_{n} is at most *n*–1 (and, in fact, equality implies that the set generates *S*_{n}).

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).

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## About Peter Cameron

I count all the things that need to be counted.

“Doing research” is my top post today, so I thought I would just add that I came up with this little trick during a sleepless night last night.

Also, I didn’t say explicitly, but Julius used the Classification of Finite Simple Groups in his proof. I would be prepared to buy a drink for anyone who can come up with a proof of the answer to Vivek Jain’s question without using CFSG. (Sorry, there are no CameronCounts T-shirts or mousemats!)

Here is a slight strengthening. We can choose the transversal so that the representative of the subgroup

His the identity element. For consider the remainingn–1 elements. If they are dependent, argue as before; if they are independent, then Whiston’s Theorem tells us that they generate the symmetric group, and we are done.I guess that most transversals generate

G(as a function of the index |G:H|=n). Any thoughts about this?