A kind of converse of Lagrange’s Theorem Let $G$ be a finite group. If $a\in G$, a consequence of Lagrange’s Theorem is that the order of $a$ divides the order of $G$. Let $p=|G|$. If $p$ is prime, it is well known that $G$ is cyclic, and therefore, there exists $a\in G$ such that the order of $a$ is $p$ and, of course, there exists $b$ in $G$ such that the order of $b$ is $1$. If $n$ is a composite number, namely $$n=p_1^{m_1}p_2^{m_2}\ldots p_r^{m_r},$$
where $p_i$ are primes, my question is this:
Is there, for each $i$, an element $a_i$ in $G$ such that the order of $a_i$ is $p_i$? Is there, for each $i$, $b_i\in G$, such that the order of $b_i$ is $p_i^{m_i}$?
 A: To add on to what Matt stated, since Sylow's theorem guarentees the existence of a p-subgroup of order $p_{i}^{m_{i}}$, we have that $G$ also contains subgroups of order $p_{i}^{k}$ for all $k \le m_{i}$
A: In addition to Cauchy's Theorem and Sylow's Theorems, there is also Hall's Theorem, which states that if $G$ is solvable, $|G|=mn$ with $\gcd(m,n)=1$, then $G$ has a subgroup of order $m$, and conversely, that if $G$ has this property, then $G$ is solvable.
So, in summary:


*

*The converse of Lagrange's Theorem ("if $a$ divides $|G|$ then $G$ has a subgroup of order $a$") is false in general; of course, this implies that the even stronger statement you were hoping for ("if $a$ divides $|G|$, then $G$ has an element of order $a$") is also false.

*However, there are a number of restrictions that make the converse true:


*

*If $a$ is prime, then the result holds; this is Cauchy's theorem: if $p$ is a prime and $p$ divides $|G|$, then $G$ has an element (and hence a subgroup) of order $p$.

*In fact, the converse of Lagrange's theorem holds for prime powers, which is one of the Sylow theorems: if $p$ is a prime, $n$ a nonnegative integer, and $p^n$ divides $|G|$, then $G$ has a subgroup of order $p^n$.

*The somewhat stronger converse to Lagrange's theorem holds for solvable groups (and only for solvable groups): a finite group $G$ is solvable if and only if for any positive integers $n$ and $m$, if $\gcd(n,m)=1$ and $|G|=nm$, then $G$ has a subgroup of order $n$.

*The full converse of Lagrange's theorem does hold for certain kinds of groups; notably, it holds for abelian groups: if $G$ is abelian, and $a$ divides the order of $G$, then $G$ has a subgroup of order $a$. (It also holds for finite nilpotent groups).

*However, even for the classes of groups where the full converse to Lagrange's theorem holds, the element-order-version does not generally hold. Certainly, it is not the case that if a prime power $p^n$ divides the order of $G$ then $G$ has an element of order $p^n$ (that would imply that every group of prime power order is cyclic; smallest counterexample is the Klein $4$-group); even if you require that the order of $G$ not be a prime power. 
A: This is Cauchy's theorem, which states that if $p$ is a prime dividing the order of a finite group $G$, then there exists an element of $G$ with order $p$.
You might also be interested in the Sylow theorems, which is another partial converse to Lagrange's theorem, which show that if $p^n$ is the highest power of a prime $p$ dividing the order of $G$, then there is a subgroup of order $p^n$.
A: The answer to the second part of your question: "Is there, for each $i$, $b_i\in G$, such that the order of $b_i$ is $p_i^{m_i}$?" is no.  For why not, I encourage you to consider groups of the form $C_p \times C_p \times \cdots \times C_p$ where $C_p$ is cyclic of order $p$.
