Group proofs with $g^N=e$. Let $G$ be a finite group.
I want to prove, that:
$$\forall g \in G, \exists N \in\Bbb N>0, g^N = e $$
as well as
$$\exists N \in\Bbb N>0, \forall g \in G, g^N = e.$$
My approach:
If $g=e$ than it is trivial:
$$g^N = e^N = e$$
But it is hard to unterstand the rest, in this case $g\neq e$.
 A: First, note that if $G$ is a finite group then for all $g \in G$ there is $n_g \in \mathbb{N}$ such that $g^{n_g}=e$. In fact, suppose not: then at some point, since $G$ is finite, you must have $g^i = g^j$ for some $i < j$. But then $g^{j-i}=e$ (multiply $i$ times left and right by $g^{-1}$).
So to each $g$ you can associate a natural number $n_g$. Now take $N=\prod_{g \in G} n_g$. For any $h \in G$, what is $h^N$? Well:
$$h^N = h^{n_h \cdot \prod_{g \in G, h \neq g} n_g} = (h^{n_h})^{\prod_{g \in G, h \neq g} n_g} = e^{\prod_{g \in G, h \neq g} n_g} = e$$
This is far from the optimal $N$ for which this holds, but it avoids using Lagrange's theorem or any other result really.
A: First part : Let $n$ be the cardinality of $G$. Consider the $n+1$ elements $e,g,g^2, ..., g^n$. You have $n+1$ elements from a set of $n$ elements, therefore two of them are equal. So there exists $k > l \in \lbrace 0, ..., n \rbrace$ such that $g^k=g^l$. Now let $N_g = k-l > 0$ : for this $N_g$, you have
$$g^{N_g}=e$$
Second part : Apply Lagrange's theorem to the subgroup generated by $g$, to show that each $N_g$ must divide the order $n$ of $G$, and therefore that $n$ works for all the $g$'s in $G$.
