Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$, where $\sigma\in S_{14}$ and $|\sigma|=28$? 
Let $\sigma\in S_{14}$ which is an even permutation of the order of $28$. Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$.

My try:
We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and $28=2\cdot2\cdot7$.
So $\sigma$ must be character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ - composition of row cycle $4$ and row cycle $7$
 because if $\sigma$ would be a character $(a_1
,a_2)(b_1b_2)(c_1...c_7)$ then $|\sigma|=2\cdot7=14$ which is contrary to the assumption.
That's why $4+7=11$ elements elements undergo nontrivial permutations and $14-11=3$ elements pass on to each other.
So $\sigma$ has a character:
$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$
Moreover we have information that $\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ is composition an even number of transpositions.
However these are my only thoughts and I don't know what to do next to come to the thesis.
EDIT:
According to the remark from @EricTowers $\sigma$ can still have a character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$ then $$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$
 A: Let $n_k$ be the number of $k$-cycles in the disjoint cycle decomposition of $\sigma$. Then


*

*$k \in \{1,2,4,7,14,28\}$

*$n_{28}=0$ because $28>14$

*$n_{14}=0$ because a $14$-cycle is not even

*$n_4 \ge 1$

*$n_7 \ge 1$

*$n_1 + 2n_2 + 4n_4 +7n_7 = 14$
The last equation has no solutions if $n_1=0$. Thus $n_1\ge1$, as required.
Actually, the only solutions are
$(n_1,n_2,n_4,n_7)=(1,1,1,1)$ and $(3,0,1,1)$.
However, since $\sigma$ is even, we must have $n_2+n_4$ even and so the only solution is $(n_1,n_2,n_4,n_7)=(1,1,1,1)$.
A: It is not required that three elements are fixed.  Consider
$$  (1\ 2\ 3\ 4\ 5\ 6\ 7)(8\ 9\ 10\ 11)(12\ 13)(14)  \text{.}  $$
If the order of the cycle is $28$, there is at least a $7$-cycle and at least a $4$-cycle, as you have shown.  Any other cycle's length must divide $7$ or $4$.  How many ways can the three elements not in those two cycles be distributed among divisor-or-$7$ or divisor-of-$4$ cycles?
A: You already have the prime factorisation of $28$. To get an element of order $28$, you need to partition $14$ into divisors of $28$ (namely, $1$, $2$, $4$, $7$, and $14$) so that their LCM is $28$.${}^\dagger$ So, what are the partitions of $14$ into those divisors, potentially including $1$, $4$, and $14$, such that the disjoint cycles of elements of $S_{14}$ form elements of order $28$ with cyclic decompositions composed of those divisors?
You'll find that you'll always need a $1$ in the cyclic decomposition. What does that imply?
You need to have at least one term of $7$ or $14$ in the partition. It should be obvious why you can't have a term $14$; can you have two terms of $7$? If the number of $7$s in the partition is odd, what does that say about the number of $1$s in the partition?${}^\dagger$
$\dagger$: I am thankful to @StevenStadnicki for the clarifying sentences provided in the comments.
