Prove or give a counterexample: If a group G has a subgroup of order n, then G has an element of order n The question is given in the title. I am unable to come up with a counterexample and I'm thinking this could apply to cyclic groups but not necessarily to general groups. Does anyone have any ideas or something to point me in the right direction?
Help would be much appreciated, thank you!
 A: Hint : think to a product $G=A \times B$ of some groups $A$ and $B$...
The elements in $G$ have to form $(x,y)$. You have
$$m \cdot (x,y) = (mx,my) =(0,0).$$ In particular, if $x$ and $y$ have order dividing $m$ (that is $mx=0=my$), then $(x,y)$ has also an order that divides $m$ (that is $m(x,y)=(0,0)$). 
If $A$ and $B$ have order $m$, then every $x \in A$ and every $y \in B$ have order that divides $m$. From what we've just said above, every element in $A$ has also order that divides $m$. In particular, no element in $G$ has order $m^2 = |G| = |A| \cdot |B|$ (if $m≥2$).

 For instance, you can take $A = B = \Bbb Z / 2\Bbb Z$, so that $m=2$. Then $G  =\Bbb Z / 2\Bbb Z \times \Bbb Z / 2\Bbb Z$ has a subgroup of order $m^2=4$ (namely itself) but has no element of order $4$. Indeed, every element in $A$ has an order $d$ that divides $2$.

A: There are $n$'s for which every group of order $n$ is cyclic, for instance $n$ prime, or $n = p q$, with $p, q$ prime, $p > q$ and $p \not\equiv 1 \pmod{q}$. (In general, all groups of order $n$ are cyclic iff $\gcd(n, \varphi(n)) = 1$ iff $n = p_{1} p_{2} \dots p_{k}$, with the $p_{i}$ primes such that $p_{1} > p_{2} > \dots p_{k}$, and $p_{i} \not \equiv 1 \pmod{p_{j}}$ for all $i, j$.)
So let $n$ be such that there is a non-cyclic group $G$ of order $n$. Such a $G$ will be your generic counterexample.
Namely, if $n$ is divisible by the square $p^{2}$ of a prime, consider first a non-cyclic group $H$ of order $p^{2}$, and $G = H \times K$, with $K$ cyclic of order $n/p^{2}$. 
If $n$ is not divisible by the square of a prime, and $n = p_{1} p_{2} \dots p_{k}$, with the $p_{i}$ primes such that $p_{1} > p_{2} > \dots p_{k}$, then one has $p_{i} \equiv 1 \pmod{p_{j}}$ for some $i > j$. Then start with a non-abelian group $H$ of order $p_{i} p_{j}$, and obtain $G = H \times K$,  with $K$ cyclic of order $n/(p_{i} p_{j})$. 
A: Look at $S_3$. It's got a subgroup (the whole group) of order 6, but no element of order 6. 
A: Let $G$ be a group of order $n$. Then:
$G$ has an element of order $n$ iff $G$ is cyclic.
Hence every non-cyclic group provides you with a counterexample.
A: The group $\;\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2\;$ of order eight has a subgroup of order four, namely $\;\{0\}\times\Bbb Z_2\times\Bbb Z_2\;$, but all its elements have order one or two.
A: Is false except for cyclic groups: If $G$ isn't cyclic then has a subgroup (the same $G$) of order $|G|$ but all the elements are of order $<|G|$.
