$(1 2)(3 4)$ does not commute with any nonidentity element of odd order in $A_5$. On Dummit's Abstract Algebra on p. 128, it says:

"It is easy to see that $(1 2)(3 4)$ ... does not commute with any non-identity element of odd order in $A_5$."

But I don't find it easy.
Any help, please?
 A: Here's a bit of reasoning that generalizes a little better than just brute force. 
This is equivalent to showing that $\tau(12)(34) \tau^{-1}\neq(12)(34)$ for the $\tau$ you specified. Fortunately this is easy to compute using the trick that $\tau(12)(34) \tau^{-1}=(\tau(1)\tau(2))(\tau(3)\tau(4))$. (My order of composition is to execute the rightmost permutation first.)
So if we tried to find a nonidentity permutation $\tau$ such that $\tau(12)(34) \tau^{-1}=(12)(34)$, it would have to move at least one of $\{1,2,3,4\}$, and $5$ is fixed. There are then these cases:


*

*The symbols $1,2$ are transposed, the symbols $3,4$ are transposed, or maybe both pairs are transposed. 

*The symbols $1,3$ are transposed (and accordingly $2,4$ are transposed), or else $1,4$ are tranposed (and accordingly $2,3$ are transposed.)
A single transposition is of even order, so the only candidates are products of two transpositions, which are all three cycles. But that means that exactly one of $\{1,2,3,4\}$ is fixed by $\tau$. How could this be? If you fix any symbol in $(12)(34)$ and want to preserve the equality above, then you automatically fix its partner. By this contradiction, we see no $\tau$ exists.
A: Assume $\tau=(12)(34)$ commutes with $\sigma$. You have two situations:
Case 1: $\sigma(5)= a  \neq 5$.
Then $\sigma(\tau(5))=\sigma(5)=a$ while  $\tau \sigma(5) = \tau(a)$. As $a \in \{1,2,3,4\}$ it follows that $\tau(a) \neq a$ which proves that
$$ \sigma(\tau(5)) \neq \tau (\sigma(5))$$
Case 2: $\sigma(5)= 5$. Then, sigma must have another fixed point $a \in \{ 1,2,3,4 \}$, which is unique. Then, as $\tau(a) \neq a$ we have 
$\sigma(\tau(a))\neq \tau (a)$ while  $\tau( \sigma(a)) = \tau(a)$.
