Question about Lagranges theorem in Group Theory

[3029298]

Homework Statement

If H is a subgroup of a finite group G, and if the order of G is m times the order of H, |G|=m|H|, adapt the proof of Lagrange's theorem to show that g^m! is an element of H for all g in G.

The Attempt at a Solution

My thoughts so far were to think that we can divide G into m pieces: H, g₁H, g₂H, ..., g_m-1H. If we now take an arbitrary g_k, it is certainly in g_kH, since H contains the identity. The question now is: what happens if we look at g_k², (g_k²)³... where do they end up? I need a hint...

 

[JSuarez]

If g is in H, there's nothing to prove; if not, consider the m (left) cosets:

$$H,gH,g^{2}H,...,g^{m-1}H$$

Now, if you apply the translation by g again, you get:

$$gH,g^{2}H,...,g^{m}H$$

But this translation is bijective, and that implies that $$g^{m}H$$ must be equal to what? And from this, what may be inferred about $$g^{m}$$ (and also $$g^{m!}$$)?