Quotient Groups

When $N$ is a normal subgroup of a group $G,$ the quotient group $G/N$ is obtained by "collapsing the elements of $N$ to the identity." More precisely, the set $G/N$ is defined as the set of equivalence classes where two elements $g,h$ are considered equivalent if the cosets $gN$ and $hN$ are the same.

By far the most well-known example is $G = \mathbb Z, N = n\mathbb Z,$ where $n$ is some positive integer and the group operation is addition. Then $G/N$ is the additive group ${\mathbb Z}_n$ of integers modulo $n.$ So the quotient group construction can be viewed as a generalization of modular arithmetic to arbitrary groups. In fact, the quotient group $G/N$ is read "$G$ mod $N.$"

The quotient $G/H$ is a well-defined set even when $H$ is not normal.

Let $G$ be a group and $H$ a subgroup. Then $G/H$ is the set of left cosets $gH = \{gh \colon h \in H\},$ as $g$ runs over the elements of $G.$

If $N$ is normal, then the set $G/N$ has a natural group structure; because $Ng_2 = g_2N$ $$(g_1N)(g_2N) = g_1g_2NN = g_1g_2N.$$ This gives a formula for multiplying cosets. Another way to express this formula is as follows:

If $N$ is a normal subgroup of $G,$ then the function $\pi \colon G \to G/N$ given by $\pi(g) = gN$ is a group homomorphism.

The definition of the quotient group uses cosets, but they are somewhat unwieldy to work with. It is often easier to denote the coset $gN$ by the notation $\overline g$; then ${\overline g_1}{\overline g_2} = {\overline{g_1g_2}}$ as expected. The important point is that this is true no matter which representatives $g_1,g_2$ are chosen: if ${\overline{g_1'}} = {\overline{g_1}}$ and ${\overline{g_2'}} = {\overline{g_2}},$ then $$g_1'g_2' = g_1 n_1 g_2 n_2 = g_1 g_2 \big((g_2^{-1} n_1 g_2) n_2\big) \in g_1g_2N,$$ so $\overline{g_1'g_2'} = \overline{g_1g_2}.$ Another way to say this is as follows: the coset containing the product of two coset representatives is independent of the choice of representatives.

This is not true if $N$ is not normal.

Let $G = S_3,$ the symmetric group on three symbols. Let $H$ be the two-element subgroup generated by the transposition $(12).$ Then $G/H$ consists of three cosets:

• $H = \{id, (12)\}$

• $(13)H = \{(13),(123)\}$

• $(23)H = \{(23),(132)\}$.

Since $H$ is not normal, it does not inherit the group structure from $G$; in other words, the product of two coset representatives will land in a coset that is not independent of the choice of representatives.

For example, taking $id \in H$ and $(13) \in (13)H,$ the product of these two representatives is $(13),$ which is in $(13)H.$ But instead taking $(12) \in H$ and $(13) \in (13)H,$ we find $(12)(13) = (132) \in (23)H$.

When $G = \mathbb Z$ (with group law given by addition) and $N = 6{\mathbb Z},$ the quotient $G/N$ is the set of cosets of $N.$ The coset representatives that are usually chosen are $0,1,2,3,4,5$. So for instance the coset $1+6{\mathbb Z}$ is abbreviated ${\overline{1}} \in {\mathbb Z}_6,$ and $${\mathbb Z}_6 = \left\{{\overline{0}}, {\overline{1}}, {\overline{2}}, {\overline{3}}, {\overline{4}}, {\overline{5}} \right\}.$$ The addition in ${\mathbb Z}_6$ is as expected: ${\overline{a}}+{\overline{b}} = {\overline{a+b}}.$ If $a+b > 6,$ subtracting $6$ will give the coset representative in the range $\{0,1,2,3,4,5\}.$ For example, $${\overline{3}}+{\overline{5}} = {\overline{8}} = {\overline{2}}.$$ Here ${\overline{8}} = {\overline{2}}$ because $8-2 \in 6{\mathbb Z},$ so $8+6{\mathbb Z} = 2+6{\mathbb Z}.$

Main article: Group isomorphism theorems

The three fundamental isomorphism theorems all involve quotient groups. The most important and basic is the first isomorphism theorem; the second and third theorems essentially follow from the first. Here are some examples of the theorem in use.

(First Isomorphism Theorem) A group homomorphism $\phi \colon G \to H$ induces an isomorphism $${\overline{\phi}} \colon G/\text{ker}(\phi) \to \text{im}(\phi)$$ defined naturally by ${\overline{\phi}}({\overline{g}}) = \phi(g).$

The complex numbers $z$ such that $|z|=1$ form a group under multiplication. Call this group $C$ (for unit circle). Show that $${\mathbb R}/{\mathbb Z} \cong C,$$ where $\mathbb R$ and $\mathbb Z$ denote the additive groups of real numbers and integers, respectively.

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Consider the function $\phi \colon {\mathbb R} \to C$ given by $\phi(r) = e^{2\pi i r}.$ Then $\phi$ is clearly surjective, because every complex number with absolute value $1$ can be written as $e^{i\theta}$ for some real number $\theta$ (by Euler's formula). The kernel of $\phi$ is the set of real numbers $r$ such that $e^{2\pi i r} = 1,$ i.e. $\cos(2\pi r) = 1$ and $\sin(2\pi r) = 0.$ This happens if and only if $r$ is an integer, so $\text{ker}(\phi) = \mathbb Z.$

The result follows directly from the first isomorphism theorem. $_\square$

Another example of the first isomorphism theorem is an appealingly nontrivial example of a non-abelian group and its quotient.

Consider the symmetric group $S_4$ on four symbols. It permutes the vertices of this tetrahedron: Disjoint pairs of edges are preserved. ^[1] There are three pairs of disjoint edges: the two purple edges, the blue/green pair, and the red/yellow pair. Any permutation of the vertices will permute the edges in such a way as to move these pairs onto each other. For instance, a transposition of the red and yellow vertices will fix the purple edge pair, but the red/yellow pair will swap places with the blue/green pair.

So any permutation in $S_4$ will have an associated permutation of these three objects (the edge pairs). This gives a function $\phi \colon S_4 \to S_3.$ It is a homomorphism (essentially tautologically, since the group operation on both sides is just composition of functions). Its kernel $V_4$ has four elements in it, consisting of the identity and the three double transpositions. (Any double transposition will fix both edges in one pair, and will swap the edges in the other two pairs. For example, swapping the yellow and red vertices and then swapping the blue and green vertices will leave the purple edge pair unchanged, but will swap the blue and green edges, and the yellow and red edges. The three pairs stay in the same place, even though the two edges in some pairs may have switched places.) And it is not hard to show that $\phi$ is surjective.

So the first isomorphism theorem gives an isomorphism ${\overline{\phi}} \colon S_4/V_4 \cong S_3.$

1. Atiliogomes, . Adjacent-vertex-distinguishing total coloring of complete graph K4. Retrieved May 25, 2013, from https://commons.wikimedia.org/wiki/File:Avd-total-coloring-of-complete-graph-K4.svg