Okay, so I went and tackled this question a different way and came up with:
Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional S must be invertible since it is injective. Thus if ST=I then TS=I.
Assume TS=I. Let v' be an element of V such that Sv' is in the null space of T. v'=Iv'=TSv'=0. Since V is finite dimensional T must be invertible since it is injective. Thus if TS=I then ST=I.
Therefore ST=I if and only if TS=I.
Does this make sense?
So could itb e changed to:
Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional T must be invertible since it is injective. Thus if ST=I then TS=I.
and also best to add:
ST = I
TSTT' = TIT' (where T' is the inverse of T)
TS = I
or am i still on the wrong track?
This just follows from the fact that composition of invertible linear transformations is associative (function composition in general is associative). If then
but by associativity
So . Multiplying on the left by on both sides yields .
The above obviously holds for any group.