Why Does \( g^{m!} \in H \) for All \( g \in G \)?

[mathmadx]

Dear all,
The question I've been struggling with is supposed to be solved using the way Lagrange's thm was proven( with number of cosets and stuff). However, it remains a mystery how to do it:

Let G be a finite group and H<G with |G|=m|H|. Proof that

g^{m!} \in H, \forall g \in G

 

[rasmhop]

Using Lagrange's theorem on your expression we get [G : H] = |G|/|H| = m so there are m cosets of H in G. Consider:
1H,gH,g^2H,\ldots,g^mH
Can they all be different? If not then we have an integer n \in \{1,2,\ldots,m\} such that g^n \in H. Now since H is a group and m!/n is an integer, then H contains:
\left(g^n\right)^{m!/n} = g^{m!}

 

[mathmadx]

Ok, thanks, however, it's still a bit blurry to me why the fact that they can't be all the same implies that there is such an integer n..

 

[rasmhop]

Ok, thanks, however, it's still a bit blurry to me why the fact that they can't be all the same implies that there is such an integer n..

Since they are not all the same we can find two that are equal. Let these be g^a H and g^b H with a,b \in \{0,1,\ldots,m\}. Assume without loss of generality that a < b, then we have:
g^b H = g^a H
which implies
g^{b-a} H = g^{a-a}H = H
which is equivalent to g^{b-a} \in H.

Since b \leq m and a is non-negative b-a\leq m and since b>a we have 0<b-a which together gives us 0 &lt; b-a \leq m so n=b-a is the number we wanted to prove existed.