TITLE: Give an example of: A group with an element A of order 3, an element B with order 4, where order of AB is less than 12
QUESTION [18 upvotes]: I'm a mathematics major studying at University as an undergrad.  This is a question on the study guide for the upcoming final in Math 344 - Group Theory:
"Give an example of a group G with an element a of order 3, an element b of order 4, where order of ab is less than 12."
My understanding is if an element a has order n, it means that if a is combined with itself n times, it results in the identity element e: a^n=e.  It also means that there is no number smaller than n where this is true for a. Two elements a and b can be combined into ab such that ab is the result of whatever operator acts on the group. Example: If the operator is addition, ab=a+b
Possible groups I've considered that don't seem to work:
-D2n, the group of symmetries of a regular n-sided polygon.  This includes rotations about the center, or flips across lines that go through the center.  It doesn't seem to work because if a rotation has order 3, and another rotation has order 4, their combination should have order 12.  All the flips or combinations of a rotation with a flip have order 2
-Quotient group Z/nZ.  Z/12Z doesn't seem to work, since {12Z+4} is order 3, {12Z+3} is order 4, but {12Z+3+12Z+4}={12Z+7}, which has order 12.  This seems to hold for other values of n
-The group of integers/reals/rationals with the addition operator, or the group of non-zero real numbers with multiplication, or the group of rationals with multiplication.  None of these seem to have elements of order 3 or 4 in the first place
These are the main groups we worked with in class.  I've searched this site and others for examples of groups I may have overlooked, with no luck.  I believe the elements I need won't be commutative - such that ab does not equal ba - but I'm not certain.
Thank you!

REPLY [6 votes]: Let $A$ and $B$ be the quaternions $i$ and $\cos(2\pi/3) + j\sin(2\pi/3)$. Then $AB = i\cos(2\pi/3)+k\sin(2\pi/3)$. Since $AB$ is a vector quaternion of unit length, it follows that $(AB)^2=-1$. So $AB$ has order $4$.
The resulting group is $\mbox{Dic}_3$. See dicyclic groups.