1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rama_ran Sep 20 '15 at 12:00
  • $\begingroup$ 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. $\endgroup$ – Groups Sep 20 '15 at 12:11
  • $\begingroup$ But only product of such $n_i$'s (though it is larger than lcm) gives us the answer. $\endgroup$ – rama_ran Sep 20 '15 at 12:16
2
$\begingroup$

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 $G$).

$\endgroup$
  • $\begingroup$ I know Lagrange's Theorem. Would you explain the answer in details $\endgroup$ – rama_ran Sep 20 '15 at 13:41
  • $\begingroup$ I editet my answer. $\endgroup$ – M.U. Sep 20 '15 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.