After I wrote my last post on the condition in a group, I had a sudden doubt concerning the case in which this arose: there we assume that we have a coset
such that
for all
. I claimed that this implies
, but really the argument I wrote just means that
: for all
, we get
, hence
. But what about the other inclusion?
I had in mind a case where the groups involved are finite, or close enough, so that the reverse inclusion is indeed obviously true. But it is amusing to see that in fact what I wrote is correct in all cases: if is a subgroup of an arbitrary group
and
satisfies
, then in fact
: taking the inverse of the inclusion gives
I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition is not, in general, equivalent with
.
(I found this explained in an exercise in Bourbaki, Algèbre, Chapitre I, p. 134, Exercice 27, or page 146 of the English edition; here is a simple case of the counterexample from that exercise: consider the group of permutations of
; consider the subgroup
which is the pointwise stabilizer of
, and the element
which is just
Then we have , because the left-hand side is the pointwise stabilizer of
which is a subset of
. But
is not equal to
, because
is the pointwise stabilizer of
, and there are elements in
which fix
but not
.)
It seems natural to ask: what exactly is the set of all
such that
is equivalent to
? This set
contains the line
for
, and also the coordinates axes
and
(since we then deal with cosets of
).
In fact, one can determine : it is the complement of the line
, for
(which corresponds to conjugation). Indeed, suppose for instance that
. Let
and
be such that
.
We can write
and
Since , and since
is the common exponent of
and
at the two extremities of the right-hand side, it follows that this right-hand side is contained in
.
A similar argument works for , using
where again it is crucial that the coefficient appears on both sides, and that it is
.
Since we already know that , and in fact that
for
(the same setting as the counterexample above works, because the
we used is an
-th power for every
), we have therefore determined the set
…
Fun!