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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.

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Thanks. It is very nice example. Do you think there is any example for unsolvable groups(Not simple)? – S. T Sep 8 '12 at 15:38
@Derek: Each sporadic group has a prime p such that the normalizer has either odd order or twice-odd order, so always has a (usually highly) even number of Sylow p-subgroups. I think Lie types in defining characteristic should always be missing part of the Weyl group in the normalizer, so always should have an even number of defining characteristic sylow subgroups. (This leaves alternating and characteristic 2 groups to check.) – Jack Schmidt Sep 8 '12 at 15:57
Just to make sure the OP is not asking an easier question: PSL(2,11) has 55 Sylow 3-subgroups, so there is at least one odd p with an odd number of Sylow p-subgroups. Derek is answering the harder question of whether a group can be such that every odd p dividing the order has an odd number of Sylow p-subgroups greater than 1. – Jack Schmidt Sep 8 '12 at 16:13
Alternating groups are easy: take the largest prime $p$ less than $n$. The cylic Sylow $p$-group will have an even number of conjugates. – user641 Sep 8 '12 at 17:30

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.

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