# If $(G, o)$ be a finite group with identity $e$, prove that there exist a +ve integer $m$ such that $a^m=e$ holds for all $a\in G$

If $(G, o)$ be a finite group with identity $e$, prove that there exist a +ve integer $m$ such that $a^m=e$ holds for all $a\in G$.

Approach:

Edit

Since $G$ is finite, then G has finite number of elements. Assume $G=\{a_1, a_2, a_3, \cdots, a_N\}$ for some positive integer $N$. For some $a_i\in G$, consider the sequence $a_i^1, a_i^2, a_i^3, \cdots$. All these elements are in $G$ (closed under binary operation). since G is finite, then there must be repetitions in the above sequence, i.e. $a^j_i=a^k_i$ for some positive integers $j$ and $k$ with $j > k$ (without any loss of generality). i.e $a_i^{j-k}=e$ i.e $a_i^{n_i}=e$.

Here the problem is proved or not? Please suggest me.

For each $a\in G$, there is $n_a$ such that $a^{n_a}=1$. If you take product of all such $n_a$'s (or LCM), what happens?
• I have edited my answer. This is the detailed answer for you Hints of the 1st part. Would you please suggest the $lcm$ issue you told in the 2nd part of your Hints. Why this 2nd part is necessary? Why only 1st part proves the problem? Please suggest me in details. – rama_ran Sep 20 '15 at 12:00
• Suppose for example, $a^{10}=e$ and $b^6=e$. Then taking the product $10.6$ we get $a^{10.6}=(a^{10})^6=e^6=e$ and similarly, $b^{10.6}=(b^{6})^{10}=e$. Thus the product $10.6=60$ when raised to both and $b$ gives us identity. But, suppose we take LCM(10,6)=30, then raising this smaller power to both $a$ and $b$ still gives identity. – Groups Sep 20 '15 at 12:11
• But only product of such $n_i$'s (though it is larger than lcm) gives us the answer. – rama_ran Sep 20 '15 at 12:16
By Lagrange's Theorem you directly obtain $a^{|G|} = 1$ for any $a \in G$, where $|G|$ is the cardinality of $G$. More precisely:
Consider the subgroup $\langle a \rangle \leq G$. Then $o(a) = |\langle a \rangle|$ ($o(a)$ is the order of $a$) and thus by Lagrange's Theorem we get that $o(a)$ divides $|G|$. Hence $a ^{|G|} = a^{o(a) \cdot n} = 1$ (where $n$ is the index of $\langle a \rangle$ in $G$).