Here’s my introduction to groups and the cyclic groups. The introduction to dihedral groups may also help.

We’ve seen the cyclic groups. These are exactly the groups that can be *generated* by one element and so in the finite case can be thought of as the rotational symmetries of an n-gon, a line, or the trivial group (with there being a group corresponding to each one of these, for any n). There is also an infinite cyclic group which was the first group that we encountered, (which can viewed as those rotational symmetries of the circle of the form x^{n}, where x is a particular rotation by an irrational amount).

In these posts -new groups out of old- I will show how we can combine groups that we have seen to produce new groups. The simplest way to do this is by the direct product.

**Direct products, a geometric approach.** Since we have seen the cyclic groups, we will work with two finite cyclic groups. Let’s denote these by C_{n} and C_{m} which can be viewed as the symmetries of an n-gon and an m-gon respectively. If it helps, keep particular values of n and m in mind. I’ll be working with n=4 and m=3 with my pictures below.

We have the rotations of the square about the red circle and the rotations of the triangle about the blue dot. At the moment we have not put these groups together in any way. Before continuing: how could you put these together?

Well…we could think about these shapes independently, such as keeping the triangle fixed and rotating the square by 0º, 90º, 180º, or 270º. Or keeping the square fixed and rotating the triangle by 0º, 120º, or 240º. We can then think of symmetries which combine these two types, for example rotating the square by 180º and the triangle by 120º. What do powers (iterations) of this symmetry look like? Before continuing: compute the order of this group element (think about what the identity element would need to be first).

So what’s the identity element here? Well, we want something that ‘does nothing’ when combined with any other symmetry. We combine the symmetries by doing the first symmetry (looking at what it does to the square and to the triangle for the example above) and then doing the second symmetry. Before continuing: write out all of the possible symmetries and write down the Cayley table (you could use x_{r°} for rotations of the square and y_{s°} for rotations of the triangle so that elements are all of the form x_{r°}y_{s°}).

A similar process can be carried out for any two groups, keeping them ‘independent’ from each other. We’ll come back to this once we’ve seen an algebraic definition of the direct product.

**A quick introduction to modular (or clock) arithmetic.** Consider an analogue clock with only an hour hand.

We can then think of clock arithmetic as what happens when we add or take away hours…for example if it were 3 o’clock (am or pm) and we waited 6 hours, it would then be 9 o’clock, but if it were 6 o’clock and we waited 7 hours it would then be 1 o’clock i.e. 3+6=9 but 6+7=1. Before continuing: write down an explanation for someone on how to do addition with ‘clock arithmetic’.

Your explanation may well be better than mine. The key points I would make would be that:

- there are only really 12 numbers here: 1,2,3,4,5,6,7,8,9,10,11,12
- any whole number is equivalent to one of these as the remainder when divided by 12
- adding a multiple of 12 therefore does not change the answer.

One can therefore think of this as identifying all of those numbers which have the same remainder when divided by 12. Thus 1=13=25=37=49=61=… in clock arithmetic. This can be generalised to work with any number (12 was only chosen since of its familiarity with telling the time). This is done by identifying all numbers which have the same remainder when divided by m and is then called arithmetic modulo m. We would also usually consider the m numbers worth working with as 0,1,…,m-2, m-1 as 0 makes more sense to use than m. I would recommend converting my 3 bullet points above to apply for this -more general- modular arithmetic. Also, I did a lesson on this which can be downloaded here and wikipedia has a good entry on this.

**Back to group theory.** You may have noticed that working with a set of m numbers {0,1,…,m-2, m-1} with addition modulo m produce a group. Before continuing: show that this a group. What will be the identity and, for each a in {0,1,…,m-2, m-1}, what’s the inverse of a? Associativity follows from addition of whole numbers being associative. Also, decide if you’ve seen such a group structure before…possibly a Cayley table helps? The answer is below.

Spoliers! So what do we know? Well, clearly 0 is our identity element as it was when we looked at before. Similarly we have, for any whole number a, that -a is its inverse. For any a between 1 and m-1 -exactly the a that we want to be working with in our modular arithmetic- we have that -a is between -m+1 and -1 and so m-a, which is equivalent to -a in this world of modular arithmetic, is an inverse of a and lies between 1 and m-1. We note that, from a Cayley table, or that these groups can all be generated by the number 1, we have that the integers modulo m with addition have the same structure as C_{m}.

**Direct products, the algebraic way.** We now have an algebraic way of writing down the cyclic groups. I’ve done this for C_{4}, the group of rotations of the square, below (remember red then blue to make purple).

0 | 1 | 2 | 3 | |

0 | 0 | 1 | 2 | 3 |

1 | 1 | 2 | 3 | 0 |

2 | 2 | 3 | 0 | 1 |

3 | 3 | 0 | 1 | 2 |

We can also make a group of order 4 (i.e. with 4 elements) by considering 4 elements, each of which has 2 ‘coordinates’ which are chosen to be 0 or 1. As sets, this is the cartesian product of {0, 1} with {0, 1}, denoted by {0, 1} × {0, 1}. This is equal to {(0,0), (0,1), (1,0), (1,1)}. But we’re working with groups! So let’s just keep {0, 1} as the integers modulo 2, i.e. the group we have called C_{2}, and denoted by the following Cayley table.

0 | 1 | |

0 | 0 | 1 |

1 | 1 | 0 |

We note that we could give two more Cayley tables the same as this one which may motivate what I’m doing. Have a think before reading on.

So? The Cayley table on the left has a redundant right coordinate and the 2nd Cayley table has a redundant left coordinate. This is just like looking at rotations of the square and triangle above. The first Cayley table relates to the rotations of one shape which ‘do nothing’ to the second coordinate whilst the second relates to the rotations of a distinct (but possibly identical) shape which do nothing to the first coordinate. We are now free to combine these groups so that we work ‘component-wise’. Have a go at drawing the Cayley table before looking below.

(0, 0) | (1, 0) | (0, 1) | (1, 1) | |

(0, 0) | (0, 0) | (1, 0) | (0, 1) | (1, 1) |

(1, 0) | (1, 0) | (0, 1) | (1, 1) | (0, 0) |

(0, 1) | (0, 1) | (1, 1) | (0, 0) | (1, 0) |

(1, 1) | (1, 1) | (0, 0) | (1, 0) | (0, 1) |

This is the direct product of C_{2} with C_{2}, denoted by C_{2} × C_{2}. This follows the set theory notation for the cartesian product. Write down/think of your own definition for the direct product as it has been introduced here. As we did in previous posts, it is worth considering if this is a group we have seen before.

**Definition.** Given two groups (G, ♣) and (H, ♥) we may define a group on the set G × H with the following binary operation ♠: given g_{1}, g_{2} in G and h_{1}, h_{2} in H let

(g_{1}, h_{1}) ♠ (g_{2}, h_{2}) := (g_{1} ♣ g_{2}, h_{1} ♥ h_{2})

and denote this by (G, ♣) × (H, ♥) or just G × H for short.

**Important:** check whether the direct product produces a group. What is the identity, what is the inverse for a general element (g, h) where g in G and h in H, and is the defined binary operation associative?

I will explain whether, for cyclic groups, these are new groups in the comments: make sure you have a good go yourself (draw plenty of Cayley tables and make sure you have a play)!

Ok! A hint. Work with the group described above (involving the square and triangle) using the algebraic approach. What could you try to use to generate this group? Either way, when will such an element succeed or fail to generate the group: does this happen for the example C_{2} × C_{2}?

Here is the answer, as promised! First of all, the group C3 x C4 was cyclic whilst C2 x C2 was not (though it can be generated by 2 elements). We therefore need a way to decide how many generators a direct product of 2 groups requires. If it requires 1 generator then it will be a cyclic group of finite order and so be a group that we have seen before. We note that we may always use two generators, (1,0) and (0,1) to generate a direct product of 2 cyclic groups. But when will one generator be sufficient? Well, the generator will need to generate all of the first factor and all of the second factor. Such an element would be (1,1). Other possibilities can be realised by replacing 1’s with -1’s, though this really just relates to choosing if the generator is a clockwise or anticlockwise rotation (and so is a choice that won’t make a difference). So the question becomes when (1,1) will generate the group, i.e. when (1,1) will have order equal to the order of the group Cm x Cn which is equal to mn. In the case C3 x C4, we note that (1,1) + (1,1) + (1,1) = (0, -1) and so has order 12 (meaning it generates the group). In the case of C2 x C2 the element (1,1) has order 2. This means that Ck x Ck cannot be cyclic. Extending this we have that (1,1) has order equal to the lowest common multiple of m and n. Thus if they are coprime (they have no common factors) then we will have that (1,1) in Cm x Cn will have order mn i.e. if m and n are coprime then Cm x Cn is the cyclic group of order mn, denoted Cmn. Thanks for reading.

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