# 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!

## 4 Answers

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 $$(ab)^3 = -1$$.

• You might want to mention explicitly that this means $ab$ has order $3$ (and, for ease of understanding, change $-1$ to $-I$ or better yet write out the matrix). – Greg Martin Dec 6 '18 at 8:26
• I thought it meant that ab has order 6 (because order is the smallest exponent to get to 1) – Pacopaco Dec 6 '18 at 15:39

Note: I misread the question as just asking for it to not be equal to 12. To be less than 12, you could probably proceed with the same strategy, but it would be a bit more complicated.

For order(ab) greater than 12, let $$G_1$$ be the free group of $$(a,b)$$, $$H_1$$ be the free group of $$(a^3,b^4)$$, $$G_2$$ be the quotient group of $$G_1/H_1$$. $$a$$ is order $$3$$ in $$G_2$$, $$b$$ is order $$4$$, and $$ab$$ has infinite order.

Put in slightly less technical terms, $$G_1$$ is the group formed from all possible sequences of $$a$$, $$b$$, $$a^{-1}$$ and $$b^{-1}$$. E.g. $$a^3ba^{-4}$$ would be one element. $$G_2$$ is this group, except that, given two sequences, if we can go from one to the other by inserting $$a^3$$, $$a^{-3}$$, $$b^4$$, and/or $$b^{-4}$$, then we consider the two sequences to be the same. In other words, we create $$G_2$$ by taking $$G_1$$ and simply defining $$a^3$$ and $$b^4$$ to be the identity.

One physical interpretation would be suppose we have two disks. The first can only be turned in increments of 120 degrees (clockwise or counterclockwise), and the second only in increments of 90 degrees. Let $$G_2$$ be the set of sequences of rotations of the two disks, where a sequence is considered different if it does the disks in different order (rotating the first disk, then the second, is different from rotating the second then the first), but the same if, while we're on a particular disk, it ends up at the same place. So rotating the first disk four time is the same as rotating the first disk once.

• +1 very nice, though probably not accessible yet for the OP. – Ethan Bolker Dec 5 '18 at 23:39
• A nice example, but the order of $ab$ is supposed to be less than 12. – Alex S Dec 6 '18 at 0:43
• Your $H_1$ is not normal in $G_1$ (finitely generated subgroup of infinite index is never normal in noncommutative free group). After taking normal closure your construction is fine, but writing blatantly wrong statements is not fine (I think). – xsnl Dec 6 '18 at 13:37