Had this question on a group theory exam the other day:
Let P be a p-Sylow-subgroup of the symmetric group S$_5$, where p represents a prime number. Can the normaliser N$_{S_5}$(P) be of odd cardinality?
I answered that it wasn't possible because, being a subgroup of S$_5$, |N$_{S_5}$(P)| has to divide 120. The only odd numbers that divide 120 are 3,5 and 15, and I proceeded to show that S$_5$ cannot have a normal subgroup of with any of those cardinalities.
Just wanted to make sure whether this reasoning doesn't have huge flaws and whether there's a faster way to prove it, since this took me a long time to do. Thanks in advance!
As you correctly pointed out in the comment, the flaw in your reasoning is that $N_G(P)$ does not have to be normal in $G$.
Since $120=2^3 \cdot 3 \cdot 5$, there are only three examples to check (well, to be pedantic: for $p$ any other prime the Sylow subgroup is trivial and so $N_G(P)=G$ which does not have odd cardinality).
For $p=2$, clearly $P \leq N_G(P)$, so there is nothing to check.
For $p=3$ you can easily find an element of even order in $C_G(P) \leq N_G(P)$, the transposition given by the two elements that do not appear in the $3$-cycle (e.g. if $P= \langle (123) \rangle$ then $(45) \in C_G(P)$).
For $p=5$, a little more insight is required. It is quite easy to see that no elements except its powers centralise a $5$-cycle, so you are looking for things in its normaliser that do not centralise it. From the $N/C$-theorem, these can be embedded in $\operatorname{Out}(P)=C_4$, so you are looking for a $4$-cycle. With a bit of trial-and-error, you can find $$(2453)(12345)(2354)=(14253)=(12345)^3$$