Let $G$ be a finite group of even order. Also in this group for every $p$ the number of Sylow $p$-subgroups is not equal to $1$. By Sylow's theorem we know that the number of Sylow $p$-subgroups in a finite group is equal to $1+pk$ for some $k$. Is true for any prime $p\neq 2$ the number of Sylow $p$-subgroups is even number? (If it is true I want to know why?)Thanks
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It is possible to construct examples of the type you want. Here is one such - you can use the same method to construct solvable examples divisible by 2 and any two other primes.
Let $H$ be the semidirect product of a group of order 7 with a cyclic group of order 6 acting faithfully. (So $H$ is a Frobenius group of order 42.) Let $N$ be a faithful irreducible module for $H$ over the field of order 3, and let $G$ be the semidirect product of $M$ with $H$.
It turns out that the only such module has dimension 6 - it is the deleted permutation module for the degree 7 permutation representation of $H$. So $G$ is a group of order $3^6.42 = 30618$. I don't know whether this is the smallest example.
The number of Sylow $p$-subgroups of $G$ for $p=2,3,7$ is respectively 189, 7 and 729.
It would be interesting to know whether there are any simple examples.
In $A_4$, the number of Sylow-3 subgroups is 4; in $C_7\rtimes C_3$ (the only non-abelian group of order 21), the number of Sylow-3 subgroups is 7.