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Mathematics LibreTexts

5.1: Examples of Quotient Groups

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 \(G\).


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