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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    $\begingroup$ I wish I knew how to prove that without using Lagrange's theorem. $\endgroup$ – lhf May 7 '12 at 12:14

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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  • $\begingroup$ thanks .. This theorem is exactly wat i need ! $\endgroup$ – AnkurVijay Mar 21 '11 at 16:12

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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  • $\begingroup$ Note that Corollaries 1 and 2 are actually equivalent. I wish I knew a proof of them that does not depend on Lagrange's Theorem. (There is a nice one in the case of abelian groups.) $\endgroup$ – lhf Mar 21 '11 at 16:20
  • $\begingroup$ @lhf: Yeah, more or less. Ok could you please cite the reference for the abelian groups case. I haven't seen it, so would be interesting. $\endgroup$ – anonymous Mar 21 '11 at 16:21
  • $\begingroup$ 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 en.wikipedia.org/wiki/… $\endgroup$ – lhf Mar 21 '11 at 16:28

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