If $G$ is a finite group of order $n$, then $n$ is the minimal such that $g^n=1$ for all $g \in G$? I know that if the order of a finite group $G$ is $n$ then $g^n=1$ for all $g\in G$. But, is $n$ the smaller integer that satisfies that property? There isn't another $m<n$ such that $g^m=1$ for all $g\in G$?
Thanks
 A: A finite group of order $n$ will have exponent $n$ if and only if its Sylow Subgroups are cyclic for every $p|n$.
Suppose $p^r$ is the highest power of $p$ which divides $n$ (and $r\ge 1)$. If the Sylow subgroups associated with $p$ are not cyclic, then there is no element with order $p^r$, and hence no element with order divisible by $p^r$ and the exponent is at most $\frac np$.
On the other hand if there are elements $a_p$ of order $p^r$ for all $p$ the exponent of the group cannot be lower than the least common multiple of their orders, which is $n$.
A: Third attempt.  In general no.  $\mathbb Z_2 \times \mathbb Z_2$ is the simplest exception.  Many such examples will exist.  I tried to generalize but I made mistakes.  I think if $n = pq$ for prime p and q. Then this will be true as some element g will have order p and only $g^{kp} = 1$ while another h will have order q and only $h^{jq} = 1$ so the only $m$ so that $h^m = e = g^m$ is $n = pq = lcm(p,q)$.
I think for any other $n = p^im; i > 0$ it will be possible to find groups of order n where $g^{pm} = e$ for all elements and thus this wouldn't be true.  (Ex. $\prod_i \mathbb Z_p \times G'$ where G' is a subgroup of order m.)
(Of course you will also be able to find groups where $n$ is the smallest such number.  [Ex. $\mathbb Z_n$])
I sure hope after being wrong twice that this one is correct....
=====
So I think in general, if $n = \prod p_i^{k_i}$ for primes $p_i$.  There will be groups where the lowest such number (called the exponent) is $\prod p_i^{j_i}$ for every possible $0 \le j_i \le k_i$.
At least, I really hope so.  I've embarrassed myself enough tonight.
