Divisor of the cardinality of a group Let G be an abelian group. Show that, if G is not cyclic, then for all $x\in G$, there is a divisor $d$ of $n = |G|$ which is strictly smaller than n satisfying $x^d=1$. 
I'm guessing that this is a consequence of Lagrange's Theorem. We can have that G is a disjoint union of left cosets that all have the same cardinality. So $|H| <  |G|$ since G is composed with more than just one left coset. By Lagrange's Theorem, we have that $|H|=d$ and then $d$ divides $n$. However the "if G is not cyclic" part is bothering me. Does the fact that G is not cyclic put a restriction?
 A: You've shown that $d$ divides $n$, but you haven't shown that $d \neq n$ (i.e., you haven't shown that $d$ is strictly smaller than $n$). 
This is where the "non-cyclic" condition on $G$ comes in. Suppose for the sake of contradiction that $H = G$, and that $H$ is the only non-trivial subgroup of $G$, and hence $|H| = d\mid n = |G|$ and $d = n$ for every non-identity element $h\in H$. Then $h^n = 1$, and no $m$, $1\lt m \lt n$ exists such that $h^m = n$. But then $h$ generates $H = G$, and hence $G$ is cyclic. Contradiction.
A: *

*Consider $E=\{ e \in \mathbb Z : x^e =1 \mbox{ for all } x\in G \} $, the set of exponents of $G$.
Then $E$ is a subgroup of $\mathbb Z$ and so $E=m\mathbb Z$.

*By Lagrange's Theorem, $n\in E$ and so $m$ divides $n$.

*$m$ is the lcm of the orders of all elements of $G$.

*If $a,b\in G$ with $r=ord(a)$ and $s=ord(b)$ and $\gcd(r,s)=1$, then $ord(ab)=rs=lcm(r,s)$.
The last two facts imply that there is an element of $G$ having order $m$.
Thus, either $G$ is cyclic or has exponent a proper divisor of $n$.
A: If G is a finite group and x is an element of G, then o(x) divides card G, so there exists d dividing card G such that x^d=1. As here G is not cyclic, d is strictly less than card G. (Remark. "abelian" is superfluous in the statement of your pb.)
