Group of order 4k+2. Prove that the following permutation is odd. 
Let $G$ be a group such that $|G|=4k+2$, and let $a \in G$ such that
  $a$ has order 2. Consider $f:G \rightarrow G$ given by $f(g)=ag$, 
  $g\in G$. Prove that $f$ is an odd permutation.

I was discussing the following problem with a friend, we sort of came up with the same answer, but we are not sure where the order of the group comes in play.
To prove that it is a permutation we wts that $f$ is bijection from $G$ to $G$, and to prove that it is odd, we said that since $f^2=g$, then it must be a transposition so it must be odd. Is the order of the group extra information or are we missing something?

I had written up a hint in the comments, I'll make that into an answer:

Try to write $f$ in cycle notation. What happens to the identity under $f$? And, pick a $g\in G$ and since $a$ is not the identity, you'll have cycles of length $2$ and there will be $\%@\%@$ of them. So... 

My definition of odd Permutation
A permutation of $G$ is said to be odd if there are odd number of transpositions in some cycle notation of $f$.  
Note: This requires some proof that if one cycle notation of $f$ has odd number of transpositions, so will any other cycle. (This can be thought of as a consequence of the fact that identity is a even transposition, which in turn will require a proof.)
Here's an expanded version:
Note that since $a$ is an element of order $2$, it cannot be the identity. Hence, the permutation is not identity on $G$. Now note that no element is fixed. Why? $ag=g \iff a=e_G$. This is another way of saying, that left multiplication is a transitive action. This is another way of saying that left multiplication is a Quasi Regular action. (Thanks to Schmidt for bringing to my notice an abuse of terminology through the comments!)
So, all elements end up in disjoint cycles of order $2$. So, the $f$ in cycle notation is a product of exactly $2n+1$ cycles!
Hence $f$ is an odd permutation. $\blacksquare$    
If it helps, we can take an example group: 


*

*In $\mathbb Z_6$. Let $a = \bar 3$. Note that, since this is an Abelian group, $f(g)= \overline{ \bar a +\bar g}$.


By working things out, we'll see that $$f \equiv (\bar 0, \bar 3)(\bar 1, \bar 4)(\bar 2, \bar 5)$$


*

*Similarly for $\mathbb Z_{10}$, and $a= \bar 5$


Note that, $$f \equiv (\bar 0, \bar 5 )(\bar 1, \bar 6)(\bar 2, \bar 7)(\bar 3, \bar 8)(\bar 4, \bar 9 )$$
