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

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!

You are right to have observed that your elements can't commute.

The symmetric group $S_4$ has elements of both kinds but no element of order $12$ so you have lots of choices.

I hope the symmetric groups soon become a go to place for examples.

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.

Consider \begin{align*} a = \pmatrix{1 & 1 \\ 0 & 1}, \;\;b = \pmatrix{0 & -1 \\ 1 & 0} \end{align*} in $SL_2(\mathbb{F}_3)$. The product \begin{align*} ab &= \pmatrix{1 & -1 \\ 1 & 0} \end{align*} has $(ab)^3 = -1$.