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# 5.1: Examples of Quotient Groups

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Now that we've learned a bit about normal subgroups and quotients, we should build more examples.

### Integers mod $$n$$, Again

Recall the group $$\mathbb{Z}_n$$. This can also be realized as the quotient group!

Let $$n\mathbb{Z}$$ denote the set of integers divisible by $$n$$: $$n\mathbb{Z}=\{\ldots, -3n, -2n, -n, 0, n, 2n, 3n, \ldots\}$$. This forms a subgroup: $$0$$ is always divisible by $$n$$, and if $$a$$ and $$b$$ are divisible by $$n$$, then so is $$a+b$$. Since every subgroup of a commutative group is a normal subgroup, we can from the quotient group $$\mathbb{Z}/\mathord n\mathbb{Z}$$.

To see this concretely, let $$n=3$$. Then the cosets of $$3\mathbb{Z}$$ are $$3\mathbb{Z}$$, $$1+3\mathbb{Z}$$, and $$2+3\mathbb{Z}$$. We can then add cosets, like so: $$(1+3\mathbb{Z}) + (2+3\mathbb{Z}) = 3+3\mathbb{Z} = 3\mathbb{Z}.$$ The last equality is true because $$3\mathbb{Z}=\{\ldots, -6, -3, 0, 3, 6, \ldots\}$$, so that $$3+3\mathbb{Z}=\{\ldots, -3, 0, 3, 6, 9, \ldots\}=3\mathbb{Z}$$.

Exercise 5.1.0

Write out addition tables for $$\mathbb{Z}/\mathord 5\mathbb{Z}$$ as a quotient group, and check that it is isomorphic to $$\mathbb{Z}_5$$ as previously defined.

### The Alternating Group

Another example is a very special subgroup of the symmetric group called the Alternating group, $$A_n$$. There are a couple different ways to interpret the alternating group, but they mainly come down to the idea of the sign of a permutation, which is always $$\pm 1$$. The set $$\{1, -1\}$$ forms a group under multiplication, isomorphic to $$\mathbb{Z}_2$$. The sign of a permutation is actually a homomorphism. There are numerous ways to compute the sign or a permutation:

1. Determinants. A permutation matrix is the matrix of the linear transformation of $$n$$-dimensional space sending the $$i$$-th coordinate vector $$e_i$$ to $$e_{\sigma(i)}$$. Such matrices have entries all equal to zero or one, with exactly one 1 in each row and each column. One can easily show that such a matrix has determinant equal to $$\pm 1$$. Since the determinant is a multiplicative function - $$\det (MN) = \det(M) \det(N)$$ - we can observe the the determinant is a homomorphism from the group of permutation matrices to the group $$\{\pm 1\}$$.

2. Count inversions. An inversion in a permutation $$\sigma$$ is a pair $$i<j$$ with $$\sigma(i)>\sigma(j)$$. For example, the permutation $$[3,1,4,2]$$ has $$\sigma(1)>\sigma(2), \sigma(1)>\sigma(3)$$ and $$\sigma(3)>\sigma(4)$$, and thus has three inversions. If there are $$i$$ inversions, then the sign of the permutation is $$(-1)^i$$.

3. Count crossings. Draw a braid notation for the permutation where no more than two lines cross at any point and no line intersects itself. Then count the number of crossings, $$c$$. Then $$s(\sigma)=(-1)^{c}$$. The alternating group is the subgroup of $$S_n$$ with $$s(\sigma)=1$$. (To prove that this method of counting works, one needs a notion of Reidemeister moves, which originally arise in the fascinating study of mathematical knots.)

Exercise 5.1.1

Find the inversion number for every permutation in $$S_4$$, and then sort the permutations by their inversion number.

Exercise 5.1.2

Show that each of the three definitions of the sign of a permutation give a homomorphism from $$S_n$$ to $$\{1, -1\}$$. (For the third definition, a sketch of a proof will suffice. Be sure to argue that different braid notations for the same permutation give the same sign, even if the total number of crossings is different.)

We call a permutation with sign $$+1$$ a positive permutation, and a permutation with sign $$-1$$ a negative permutation.

Exercise 5.1.3

Show that there are $$\frac{n!}{2}$$ positive permutations in $$S_n$$.

Now we can define the alternating group.

Definition 5.1.4: Alternating Groups

The alternating group $$A_n$$ is the kernel of the homomorphism $$s: S_n\rightarrow \mathbb{Z}_2$$. Equivalently, $$A_n$$ is the subgroup of all positive permutations in $$S_n$$.

Exercise 5.1.5

Write out all elements $$A_4$$ as a subgroup of $$S_4$$. Find a nice generating set for $$A_4$$ and make a Cayley graph.

In fact, the alternating group has exactly two cosets. The quotient group $$S_n/\mathord A_n$$ is then isomorphic to $$\mathbb{Z}_2$$. Figure 5.1.2: Quotient of $$S_3$$ by $$A_3$$.

### Quotients of Products

Amongst the natural numbers, if we do something like $$(m*n)/m$$, we expect to get $$n$$ back again. Is this also true for products and quotients of groups?

If we consider a product group $$G\times H$$ with identity $$(1, 1)$$, we can take the subgroup given by elements $$(1, h)$$ for $$h\in H$$. Remembering the injection map $$\iota: H\rightarrow G\times H$$, we notice that this subgroup is just $$\iota(H)$$. The cosets of $$\iota(H)$$ are given by $$(g, 1)\iota(H)$$, for $$g\in G$$. Then the quotient $$G\times H/\mathord \iota(H)$$ is isomorphic to $$G$$.

## Contributors

• Tom Denton (Fields Institute/York University in Toronto)