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How does one prove that:

for all $a\in G$, where $G$ is a group (not necessarily abelian) $a^{|G|} = 1_G$.

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    [I wish I knew](http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory) how to prove that *without* using Lagrange's theorem.2012-05-07

2 Answers 2

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First, your restrict yourself to finite groups (otherwise, the statement does not make sense).

Then, you apply Lagrange's Theorem to the subgroup $\langle a \rangle$ to conclude that the order of $a$ divides $|G|$.

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    thanks .. This theorem is exactly wat i need !2011-03-21
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This is a standard fact, which i think is proved in All abstract algebra books. Here is the proof, given in the book by Herstein. See Corollary 2.

Corollary 1: If $G$ is a finite group, and $a \in G$, then $o(a) \mid o(G)$.

Proof. Consider the cyclic subgroup generated by $a$, consisting of $a,a^{2},a^{3},\cdots,$. Since $a^{o(a)}=e$, therefore, this subgroup has atmost $o(a)$ elements. If it has fewer elements, then $a^{i}=a^{j}$, for some integers $0 \leq i < j < o(a)$. Then $a^{j-i}=e$, yet $0< j-i < o(a)$, which contradicts the meaning of $o(a)$. Thus the cyclic subgroup generated by $a$ has $o(a)$ elements, and hence by Lagrange's theorem $o(a) \mid o(G)$.

Corollary 2: If $G$ is a finite group and $a \in G$, then $a^{o(G)}=e$.

Proof. By Corollary 1, we have $o(a) \mid o(G)$ which implies $o(G)=k \cdot o(a)$, therefore $a^{o(G)}=a^{k \cdot o(a)} = (a^{o(a)})^{k} = e^{k}=e$.

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    It's a proof that is frequently used to prove Fermat's little theorem in number theory books: consider the map $x \mapsto ax$. This is a bijection. Hence $g_1 \cdots g_n = (a g_1) \cdots (a g_n) = a^n g_1 \cdots g_n$. Now cancel $g_1 \cdots g_n$. See for instance http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem#Second_method2011-03-21