Finite Group Property Proof

Question

I'm trying to prove that for every finite group $G$ there is a $n$ such that $g^{n}=1$ $\forall g \in G$.

Any ideas?

Answer 1

First, I assume you know that every subgroup $H$ of a finite group $G$ has order dividing the order of $G$. (This is proved by showing the right cosets of $H$ are a partition of $G$.)

Next, I assume you know that if $g \in G$ with $G$ finite, then $\exists m ~g^m=1$. (This is proved by noting that the set of $\{g^k~|~ k \in \Bbb N \}$ must have at least one repeated element.)

Finally, you note that $H=\{g^k ~|~k \in \Bbb N\}$ is a subgroup of $G$, so the least $k$ such that $g^k=1$ must divide $|G|$. From this fact, it follows that $g^{|G|}=1$.

Answer 2

Call $G =\{g_1,g_2, \cdots , g_k\}$. Let $\vert g_i \vert= n_i$ for $ i= 1,2,\cdots, k$. Then the integer $n=\prod_{i=1}^k n_i$ fulfills the requirement