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Let $G$ be a finitely generated group with generators $X=\{g_1,g_2,\dots,g_m\}$ and $n$ be a positive integer such that the $n$-th power of every element in $X$ is the identity.

Is it true that $\mathrm{exp}(G) | n$? Clearly for abelian groups it is true. I think that it can develop to other specific groups.

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No - there exist finitely generated torsion groups with unbounded exponent. –  user1729 Oct 30 '12 at 13:49
user1729 in this question G is not torison group and only generators have finite order. –  mojtaba farazi Oct 30 '12 at 13:55
Then you should say so in your question! So, can I ask you what you have tried? Surely this is not difficult - this is just the definition of exponent! –  user1729 Oct 30 '12 at 14:07
Just curious why you "undid" the edits that were made to improve formatting (using LaTeX)? The added "if No" is fine, but you deliberately "undid" exactly what was done to format your question. –  amWhy Oct 30 '12 at 14:19
@mojtabafarazi I corrected your copious English grammar and spelling mistakes as well as adding missing Latex. If you undo the changes bad things will happen. –  Matt N. Nov 21 '12 at 22:23
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2 Answers

No - there exist finitely generated torsion groups with unbounded exponent.

The Burnside problem asks if there exists a finitely-generated torsion group. Golod and Shafarevich proved that there does exist such a group, but the group they construct has elements of unbounded order. Novikov and Adian proved that there exists such a group of bounded order, while Ol'shanskii proved that there exists groups where every proper, non-trivial subgroup has order $p$ for some fixed prime $p>>1$ (such groups are called Tarski Monsters).

The links are all from wikipedia, as the original papers are all in Russian. Ol'shanskii wrote a book containing his proof, which you can try and read (although it is pretty close to unreadable...). It is called "Geometry of Defining Relations in Groups".

EDIT: I should say that we don't have to reach as far as infinite groups. For example, the group of symmetries of an icosahedron can be generated by and element $a$ of order two and an element $b$ of order three. However, $ab$ has order five. Of course, then one can take $a, b, ab$ as a generating set, or more generally the whole group as a generating set, to side-step this issue. One cannot do this with the infinite torsion groups I mentioned.

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+1. Is $D_{\infty}$ violets the question? –  B. S. Oct 30 '12 at 14:13
No...all known examples are infinitely presented (and complicated!). My answer is talking about groups where every element has finite order. However, it is not necessarily the case that this order is bounded. Thus, we talk about "bounded exponent" and "unbounded exponent". –  user1729 Oct 30 '12 at 14:16
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An easy counterexample is given by any symmetric group of degree $\geq 3$.

Each symmetric group is generated by the set of its transpositions, which gives $n = 2$. However starting with $S_3$, the exponent of the symmetric group is clearly not a divisor of $2$.

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