# Group theory, Lagrange theorem

1. Oct 25, 2009

### mathmadx

Dear all,
The question Ive 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

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

2. Oct 25, 2009

### 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:
$$\left(g^n\right)^{m!/n} = g^{m!}$$

3. Oct 25, 2009

### 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..

4. Oct 25, 2009

### rasmhop

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 $0 < b-a \leq m$ so n=b-a is the number we wanted to prove existed.

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