This is a really old riddle. The strategy is:
Step 1: divide the coins into 2 groups (each of 6 members) and weigh them against each other, you should know then which group of 6 coins contains the fake one.
Step 2: take the group of 6 coins containing the fake one, and divide it into two groups each with 3 coins, then weigh them against each other.
Step 3: you are left with 3 coins where one of them is fake. Weigh any two of them against each other.
Step 4: Conclude.
This is the easy version. The hard version is where the fake coin is unknown whether it weighs more or less than the rest (and for this version, our strategy here doesn't work).