Does a group of order 16 necessarily contain an element of order 4? I think the definitions are as follows: 
Order of a Group: Cardinality, i.e., the number of elements in its set.
Order of an Element: if $g$ is in $G$, then the order, $n$ would mean that $g^n = e$ (identity) 
But if I use these definitions, the question does not make much sense to me. 
Can someone please help me out?
 A: Upgrading my comment to an answer as requested. I've added a lot of detail as the OP seems to be unfamiliar with this group.
Let $\mathbb Z_2$ denote the additive group of integers modulo $2$, consisting of the elements $0$ and $1$ with the rules
$$\begin{aligned}
0+0 & = 0 \\
0+1 & = 1 \\
1+0 & = 1 \\
1+1 & = 0 \\
\end{aligned}$$
Let $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ denote the direct product of four copies of $\mathbb Z_2$. This is the group whose elements are tuples of the form $(a_1, a_2, a_3, a_4)$, where each $a_i$ is $0$ or $1$, where addition is pointwise, meaning that $(a_1,a_2,a_3,a_4) + (b_1,b_2,b_3,b_4) = (a_1+b_1, a_2+b_2, a_3+b_3, a_4+b_4)$, where each $a_i + b_i$ is addition modulo $2$.
It's easy to check that $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ is a group. It contains $2^4 = 16$ elements, since there are two possibilities for each of the four slots in the tuple $(a_1, a_2, a_3, a_4)$.
Finally, note that for each $(a_1,a_2,a_3,a_4) \in \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$, we have 
$$(a_1,a_2,a_3,a_4) + (a_1,a_2,a_3,a_4) = (a_1+a_1, a_2+a_2,a_3+a_3,a_4+a_4) = (0,0,0,0)$$
since $a_i + a_i = 0$ regardless of whether $a_i = 0$ or $a_i = 1$. This shows that the order of each element is at most $2$ (specifically, the order of the identity $(0,0,0,0)$ is $1$, and the order of every other element is $2$). In particular, there is no element of order $4$.
