On Modular Arithmetic we squashed all of $\mathbb{Z}$ down to $n$ residue classes and then did arithmetic with the classes themselves. This page does the same thing to any group: chop it into cosets, then multiply the cosets. Sometimes that works and sometimes it falls apart, and normal subgroups are exactly what decides which. As on Symmetric Groups, products of permutations are composed right to left.
Suppose $G$ is a group and $H \le G$.
In the proof of Lagrange's theorem, the left cosets were the equivalence classes of $g \sim g' \iff g = g'h$ for some $h \in H$. Equivalence classes always form a partition.
It follows from our proof of Lagrange's theorem that the collection of left cosets $\{gH : g \in G\}$ is a partition of $G$. (Recall $gH = \{gh : h \in H\}$.) So it makes sense to collect the cosets into a set of their own.
The (left) quotient space of $G$ modulo $H$ is $$G/H = \{gH : g \in G\}.$$
Each element of $G/H$ is a whole coset, a subset of $G$. Swap left for right and you get the mirror image.
$H \backslash G$ uses the same backslash as set difference. Somebody looked at this notation and decided it wasn't confusing enough yet.
The right quotient space of $G$ modulo $H$ is $H \backslash G = \{Hg : g \in G\}$.
Counting the cosets turns out to be useful enough to get its own name.
The index of $H$ in $G$, denoted $|G : H|$, is $|G : H| = |G/H|$.
If $|G| \lt \infty$ then, again from our proof of Lagrange's theorem, $$|G : H| = \frac{|G|}{|H|}.$$
That's because the proof showed $|G| = k|H|$, where $k$ is the number of cosets. Now for the question that drives the rest of the page.
We're dividing a group by a group now. Dividing by zero has to be next on the list.
Is there a natural way to use the binary operation on $G$ to turn $G/H$ into a group?
The obvious guess is "multiply two cosets by multiplying their representatives." Let's try it on two groups. In the first one it works perfectly.
You've already proved this example works: on Modular Arithmetic, addition of residue classes was shown to be well-defined. $\mathbb{Z}/n\mathbb{Z}$ was a quotient group all along.
$G = \mathbb{Z}$, $H = n\mathbb{Z}$ ($n \in \mathbb{N}$). What is $G/H$, and can we add its elements?
$$\begin{align*} G/H &= \{a + n\mathbb{Z} : a \in \mathbb{Z}\} \\ &= \{0 + n\mathbb{Z}, 1 + n\mathbb{Z}, \dots, (n-1) + n\mathbb{Z}\} \\ &= \{\overline{0}, \overline{1}, \dots, \overline{n-1}\} \end{align*}$$
(These are the residue classes mod $n$ from Logic & Equivalence Classes.) On Modular Arithmetic we saw that the rule $\overline{a} + \overline{b} = \overline{a + b}$, $a, b \in \mathbb{Z}$, is a binary operation on $G/H$ which turns this quotient space into a group.
Key fact in this example: the binary operation above is well-defined. That is, it doesn't depend on the choices of representatives for the cosets.
Now the same idea in $S_3$, where it goes wrong.
$G = S_3$, $H = \langle (1 \ 2) \rangle = \{e, (1 \ 2)\}$. List $G/H$, then try to multiply cosets by the "rule" $(g_1 H)(g_2 H) = (g_1 g_2)H$.
$$\begin{align*} H_0 &= eH = \{e, (1 \ 2)\} = (1 \ 2)H \\ H_1 &= (2 \ 3)H = \{(2 \ 3), (1 \ 3 \ 2)\} = (1 \ 3 \ 2)H \\ H_2 &= (1 \ 3)H = \{(1 \ 3), (1 \ 2 \ 3)\} = (1 \ 2 \ 3)H \end{align*}$$
(Scratch work: $(2 \ 3)(1 \ 2) = (1 \ 3 \ 2)$ and $(1 \ 3)(1 \ 2) = (1 \ 2 \ 3)$.) So $G/H = \{H_0, H_1, H_2\}$.
Problem: this "rule" is not well-defined. For example, $H_0 = eH = (1 \ 2)H$ and $H_1 = (2 \ 3)H$, but
$$(e(2 \ 3))H = (2 \ 3)H = H_1, \qquad \text{and} \qquad ((1 \ 2)(2 \ 3))H = (1 \ 2 \ 3)H = H_2.$$
Same two cosets, two different answers.
So the coset $H_0 H_1$ has no single answer: it depends on which names you used for $H_0$ and $H_1$. Hunt for that kind of clash yourself below, and then try a subgroup where you can't find one.
Tap any element on the left as $g_1$ and any element on the right as $g_2$. The product $g_1 g_2$ lights up the coset it lands in. The scoreboard remembers every coset each pair has landed in.
In $S_3$ with $\langle (1 \ 2) \rangle$, clashes are easy to find. With $A_3$, or with $\langle r^2 \rangle$ in $D_8$, every representative agrees. The difference between those subgroups is a word from the previous page.
The Sneak Peek on Normal Subgroups promised that normal subgroups are precisely what's needed to define quotient groups. Here's the payoff.
Here is the main theorem of the page. It says that well-definedness and normality are the same condition.
Suppose $G$ is a group and $H \le G$. The rule $$(*) \quad (g_1 H)(g_2 H) = (g_1 g_2)H, \quad \forall g_1 H, g_2 H \in G/H$$ is a well-defined binary operation on $G/H$ if and only if $H \unlhd G$.
When $H \unlhd G$, $G/H$ together with this binary operation is a group, called the quotient group of $G$ modulo $H$.
There are really three things to prove: well-defined implies normal, normal implies well-defined, and well-defined implies "it's a group." The first two are the heart of it.
Proof. ($\Rightarrow$) Suppose $(*)$ is well-defined. Want to show: $\forall g \in G, h \in H$, we have $ghg^{-1} \in H$ (so $gHg^{-1} \subseteq H$).
Note that $\forall h \in H$, $hH = H = eH$. Then, $\forall g \in G$,
$$(hg^{-1})H = (hH)(g^{-1}H) = (eH)(g^{-1}H) = (eg^{-1})H = g^{-1}H.$$
Therefore, since $(*)$ is well-defined, $hg^{-1} \in g^{-1}H \Rightarrow \exists h' \in H$ s.t. $hg^{-1} = g^{-1}h' \Rightarrow ghg^{-1} = h' \in H$. Conclusion: $H \unlhd G$.
($\Leftarrow$) Suppose $H \unlhd G$. Want to show: if $g_1 H = g_1' H$ and $g_2 H = g_2' H$ then $(g_1 g_2)H = (g_1' g_2')H$.
Note that $g_i H = g_i' H \Rightarrow g_i' \in g_i H \Rightarrow g_i' = g_i h_i$ for some $h_i \in H$ ($i = 1, 2$). Then
$$\begin{align*} g_1' g_2' &= g_1 h_1 g_2 h_2 = g_1 (g_2 g_2^{-1}) h_1 g_2 h_2 \\ &= (g_1 g_2)(g_2^{-1} h_1 g_2) h_2 = (g_1 g_2) h' h_2 \in (g_1 g_2)H \quad (\text{for some } h' \in H), \end{align*}$$
where $h' = g_2^{-1} h_1 g_2 \in H$ because $H$ is normal. Therefore $g_1' g_2' \in (g_1 g_2)H \Rightarrow (g_1' g_2')H \cap (g_1 g_2)H \neq \emptyset \Rightarrow (g_1' g_2')H = (g_1 g_2)H$.
It's a group. Finally, if $(*)$ is well-defined then it turns $G/H$ into a group:
Notice where normality got used in ($\Leftarrow$): we had to slide $h_1$ past $g_2$, and conjugating $h_1$ by $g_2^{-1}$ is exactly what keeps it inside $H$. With the theorem in hand, Example 2 takes one line.
So the elements of this group are sets of elements of another group. Can I take a quotient of a quotient? (Yes, actually. Nobody can stop me.)
Example 2, revisited. $G = S_3$, $H = \langle (1 \ 2) \rangle = \{e, (1 \ 2)\}$. Why did the rule fail?
Note that $(1 \ 3)(1 \ 2)(1 \ 3)^{-1} = (2 \ 3) \notin H \Rightarrow H \not\unlhd G$.
Therefore, by the theorem, multiplication of cosets in $G/H$, as described there, is not well-defined.
The theorem also tells you how to manufacture a clash in general: find a conjugate $ghg^{-1}$ that escapes $H$, and the proof of ($\Rightarrow$) hands you two names for the same coset that multiply to different answers.
Let $H = \langle (1 \ 3) \rangle = \{e, (1 \ 3)\} \le S_3$. Find explicit representatives showing that $(*)$ is not well-defined on $S_3/H$: two names for one coset which, multiplied by the same $g_2$, land in different cosets.
$eH = (1 \ 3)H = H$, so $e$ and $(1 \ 3)$ name the same coset. Take $g_2 = (1 \ 2)$. Composing right to left,
$$e(1 \ 2) = (1 \ 2), \qquad (1 \ 3)(1 \ 2) = (1 \ 2 \ 3).$$
Now $(1 \ 2)H = \{(1 \ 2), (1 \ 2)(1 \ 3)\} = \{(1 \ 2), (1 \ 3 \ 2)\}$, which does not contain $(1 \ 2 \ 3)$. So $(e(1 \ 2))H \neq ((1 \ 3)(1 \ 2))H$: the same pair of cosets $H$, $(1 \ 2)H$ produces two different answers. (This matches the theorem: $H \not\unlhd S_3$, as shown on Normal Subgroups.)
Once $H$ is normal, computing in $G/H$ is mostly computing in $G$ and then asking which coset you landed in. These five facts make that precise.
If $H \unlhd G$ then $G/H$ is a group and:
Fact (iii): every element of a coset gets to be its name tag. Very democratic.
Facts (i) and (ii) came out of the proof of the theorem. The other three are short.
Proof of (iii). The left cosets are the equivalence classes of the relation from the proof of Lagrange's theorem. If $g' \in gH$, then $g'$ lies in the class $gH$, so its own class $g'H$ is the same class. Conversely, if $gH = g'H$ then $g' = g'e \in g'H = gH$.
Proof of (iv). For $n \ge 0$, induct on $n$: $(gH)^0 = eH = g^0H$, and if $(gH)^n = g^nH$ then $(gH)^{n+1} = (gH)^n(gH) = (g^nH)(gH) = g^{n+1}H$. For $n \lt 0$, use (ii): $(gH)^n = ((gH)^{-1})^{-n} = (g^{-1}H)^{-n} = (g^{-1})^{-n}H = g^nH$.
Proof of (v). By (iv), $(gH)^k = g^kH = eH$. So some positive power of $gH$ is the identity, and $gH$ has a finite order $m$. Since $(gH)^k = (gH)^0$, part (ii) of the theorem on orders from Subgroups gives $k \equiv 0 \pmod m$. That is, $m \mid k$. $\blacksquare$
Fact (v) only says "divides." The order of $gH$ can be all of $k$, or it can drop: the powers $g, g^2, g^3, \dots$ might wander into $H$ before they get back to $e$. Watch the two walks side by side.
Pick a group, a normal subgroup $H$, and an element $g$. Each step multiplies by $g$: the left ring walks through $g^n$ in $G$, and the right ring walks through its shadow $g^nH$ in $G/H$.
Fact (iv) also makes short work of cyclic groups.
Suppose $G = \langle g \rangle$ is cyclic and $H \le G$. Explain why $H \unlhd G$, and prove that $G/H = \langle gH \rangle$. (So every quotient of a cyclic group is cyclic.)
Proof. Cyclic groups are Abelian, and in an Abelian group $ghg^{-1} = hgg^{-1} = h \in H$, so every subgroup is normal. Now let $aH \in G/H$. Since $G = \langle g \rangle$, $a = g^n$ for some $n \in \mathbb{Z}$, and by Basic Fact (iv), $$aH = g^nH = (gH)^n \in \langle gH \rangle.$$ So $G/H \subseteq \langle gH \rangle$, and the reverse inclusion is automatic. Therefore $G/H = \langle gH \rangle$. $\blacksquare$
For example, $\mathbb{Z}/n\mathbb{Z} = \langle \overline{1} \rangle$, because $\mathbb{Z} = \langle 1 \rangle$.
The usual job: given $G$ and a normal $H$, figure out which group $G/H$ "is." The index tells you how many elements $G/H$ has, and then orders of elements usually settle it.
On Groups of Order 4–8, every group of order $4$ turned out to be $C_4$ or $V_4$, and you tell them apart by counting elements of order $2$.
The next two examples use the same group and two different subgroups of order $4$. Both quotients have order $4$, but they are not the same group.
$G = C_4 \times C_4 = \langle x \rangle \times \langle y \rangle = \{(x^i, y^j) : 0 \le i, j \le 3\}$ and $H = \langle (x^2, e), (e, y^2) \rangle$. Identify $G/H$.
$H \unlhd G$, since $G$ is Abelian, so every subgroup is normal. Also $$H = \{(x^2, e)^i (e, y^2)^j : i, j \in \mathbb{Z}\} = \{(e, e), (x^2, e), (e, y^2), (x^2, y^2)\}.$$
$|H| = 4 \Rightarrow |G : H| = |G/H| = \frac{|G|}{|H|} = 4 \Rightarrow G/H \cong C_4$ or $V_4$. Listing the cosets, $G/H = \{(e, e)H, (x, e)H, (e, y)H, (x, y)H\}$:
$$\begin{align*} (e, e)H &= \{(e, e), (x^2, e), (e, y^2), (x^2, y^2)\} \\ (x, e)H &= \{(x, e), (x^3, e), (x, y^2), (x^3, y^2)\} \\ (e, y)H &= \{(e, y), (x^2, y), (e, y^3), (x^2, y^3)\} \\ (x, y)H &= \{(x, y), (x^3, y), (x, y^3), (x^3, y^3)\} \end{align*}$$
Recall: $C_4 = \langle x \rangle = \{e, x, x^2, x^3\}$ has 1 element of order 1, 1 element of order 2, and 2 elements of order 4. $V_4 = \langle a, b \mid a^2 = b^2 = e, ab = ba \rangle = \{e, a, b, ab\}$ has 1 element of order 1 and 3 elements of order 2. Now note:
$$\begin{align*} ((x, e)H)^2 &= (x^2, e)H = (e, e)H \quad ((x^2, e) \in (e, e)H) \\ ((e, y)H)^2 &= (e, y^2)H = (e, e)H \quad ((e, y^2) \in (e, e)H) \\ ((x, y)H)^2 &= (x^2, y^2)H = (e, e)H \quad ((x^2, y^2) \in (e, e)H) \end{align*}$$
Therefore all non-identity elements in $G/H$ have order $2$, so $G/H \cong V_4$.
Same group, a "tilted" subgroup this time.
$G = C_4 \times C_4$ as before, and $H = \langle (x, y^3) \rangle$. Identify $G/H$.
$H \unlhd G$ (again, $G$ is Abelian), and $H = \{(x, y^3)^i : i \in \mathbb{Z}\} = \{(e, e), (x, y^3), (x^2, y^2), (x^3, y)\}$.
$|H| = 4 \Rightarrow |G : H| = |G/H| = \frac{|G|}{|H|} = 4 \Rightarrow G/H \cong C_4$ or $V_4$. This time $G/H = \{(e, e)H, (x, e)H, (x, y)H, (x, y^2)H\}$:
$$\begin{align*} (e, e)H &= \{(e, e), (x, y^3), (x^2, y^2), (x^3, y)\} \\ (x, e)H &= \{(x, e), (x^2, y^3), (x^3, y^2), (e, y)\} \\ (x, y)H &= \{(x, y), (x^2, e), (x^3, y^3), (e, y^2)\} \\ (x, y^2)H &= \{(x, y^2), (x^2, y), (x^3, e), (e, y^3)\} \end{align*}$$
Note: $(x, e)$ has order $4$ in $G$, and $((x, e)H)^2 = (x^2, e)H = (x, y)H \neq (e, e)H$. Therefore $(x, e)H \in G/H$ has order dividing $4$, but it does not have order $2$ or $1$, so it has order $4$.
Conclusion: $G/H \cong C_4$ and $G/H = \langle (x, e)H \rangle$.
Here are both partitions of $C_4 \times C_4$ as pictures. Column $i$ and row $j$ hold the element $(x^i, y^j)$, and squares in the same coset share a color. Example 4's cosets form a plaid, and Example 5's form diagonal stripes.
The stripes are why 3b gives $C_4$. Stepping one square to the right moves you to the next stripe, so four steps visit all four stripes before you're back where you started. In the plaid, two steps to the right always return you to your starting color.
Example 6 below finds $S_n/A_n \cong C_2$, and that's no accident. $A_n$ is the kernel of the sign map $S_n \to \{\pm 1\}$, and on Isomorphism Theorems you'll see that a group modulo the kernel of a homomorphism is always isomorphic to the image.
Next, the alternating group, which was proved normal on Normal Subgroups by the index-$2$ argument. Here is a direct check.
$G = S_n$, $H = A_n$ ($n \ge 2$). Identify $G/H$.
In words, $S_n/A_n$ only remembers one thing about a permutation: even or odd. Next, a non-Abelian group with an Abelian quotient.
$G = Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$, $H = \langle -1 \rangle = \{\pm 1\}$. Identify $G/H$.
Note: $(iH)^2 = (jH)^2 = (kH)^2 = (-1)H = H \Rightarrow iH, jH$, and $kH$ have order $2$ in $G/H \Rightarrow G/H \cong V_4$.
Modding out by $\{\pm 1\}$ forgets signs, and once signs are gone, $ij = k$ and $ji = -k$ become the same coset. That is why the quotient comes out Abelian. One last example, with a quotient of order $6$.
$G = S_4 = \langle (1 \ 2), (1 \ 2 \ 3 \ 4) \rangle$ and $$H = \langle (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4) \rangle = \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}.$$ (Note $((1 \ 2)(3 \ 4))((1 \ 3)(2 \ 4)) = (1 \ 4)(2 \ 3) = ((1 \ 3)(2 \ 4))((1 \ 2)(3 \ 4))$.) Identify $G/H$.
Note that
$$\begin{align*} ((1 \ 2 \ 3)H)((1 \ 2)H) &= ((1 \ 2 \ 3)(1 \ 2))H = (1 \ 3)H \\ &= \{(1 \ 3), (1 \ 3)((1 \ 2)(3 \ 4)), (1 \ 3)((1 \ 3)(2 \ 4)), (1 \ 3)((1 \ 4)(2 \ 3))\} \\ &= \{(1 \ 3), (1 \ 2 \ 3 \ 4), (2 \ 4), (1 \ 4 \ 3 \ 2)\}, \end{align*}$$
but $((1 \ 2)H)((1 \ 2 \ 3)H) = ((1 \ 2)(1 \ 2 \ 3))H = (2 \ 3)H \neq ((1 \ 2 \ 3)H)((1 \ 2)H)$ since $(2 \ 3) \notin ((1 \ 2 \ 3)H)((1 \ 2)H)$.
Therefore $G/H$ is non-Abelian, so $G/H \cong S_3$.
Every one of these examples boils down to the same picture: write out the multiplication table of $G$, sort its rows and columns by coset, and each block of the table turns out to be a single coset. Collapse each block to one cell and you have the table of $G/H$. Try it, including once with a subgroup that isn't normal.
Pick a group and a subgroup. Rows and columns of the multiplication table are sorted by coset, and each cell is shaded by the coset its product lands in. Press COLLAPSE to shrink every block to a single cell.
When $H$ is not normal, some block holds more than one coset, so there is no single answer to put in the collapsed cell. That's the theorem as a picture. Now a few to try on your own.
Let $G = D_8 = \langle r, s \mid r^4 = s^2 = e, rs = sr^{-1} \rangle$ and $H = \langle r^2 \rangle = \{e, r^2\}$. Explain why $H \unlhd G$, list the elements of $G/H$, and identify $G/H$.
$H = Z(D_8)$ (found on Normal Subgroups), and the center of any group is normal, so $H \unlhd G$. $|G : H| = 8/2 = 4$, so $G/H \cong C_4$ or $V_4$. The cosets are
$$H = \{e, r^2\}, \quad rH = \{r, r^3\}, \quad sH = \{s, sr^2\}, \quad srH = \{sr, sr^3\}.$$
By Basic Fact (iv), $(rH)^2 = r^2H = H$, $(sH)^2 = s^2H = H$, and $(srH)^2 = (sr)^2H = eH = H$, because $sr$ is a reflection: $(sr)^2 = s(rs)r = s(sr^{-1})r = e$. All three non-identity elements have order $2$, so $G/H \cong V_4$.
Example 7 showed a non-Abelian group with an Abelian quotient. The next exercise shows that commutativity always passes down to quotients, but can't be recovered from them.
Prove that if $G$ is Abelian and $H \le G$, then $G/H$ is Abelian. Is the converse true: if $G/H$ is Abelian for some $H \unlhd G$, must $G$ be Abelian?
Proof. Every subgroup of an Abelian group is normal, so $G/H$ is a group. For $g_1H, g_2H \in G/H$, $$(g_1H)(g_2H) = (g_1g_2)H = (g_2g_1)H = (g_2H)(g_1H).$$ $\blacksquare$
The converse is false. In Example 7, $Q_8/\{\pm 1\} \cong V_4$ is Abelian, but $Q_8$ is not ($ij = k \neq -k = ji$).
The center $Z(G)$ from Normal Subgroups is always normal, so $G/Z(G)$ is always a group. It can never be a nontrivial cyclic group, and here's why.
Prove that if $G/Z(G)$ is cyclic, then $G$ is Abelian.
Proof. Write $Z = Z(G)$, and suppose $G/Z = \langle gZ \rangle$. Let $a, b \in G$. Then $aZ = (gZ)^m = g^mZ$ and $bZ = (gZ)^n = g^nZ$ for some $m, n \in \mathbb{Z}$ (Basic Fact iv). By Basic Fact (iii), $a \in g^mZ$ and $b \in g^nZ$, so $$a = g^m z_1, \qquad b = g^n z_2 \qquad \text{for some } z_1, z_2 \in Z.$$
Elements of $Z$ commute with everything, and powers of $g$ commute with each other, so
$$ab = g^m z_1 g^n z_2 = g^m g^n z_1 z_2 = g^{m+n} z_1 z_2, \qquad ba = g^n z_2 g^m z_1 = g^n g^m z_2 z_1 = g^{m+n} z_1 z_2.$$
So $ab = ba$ for all $a, b \in G$, and $G$ is Abelian. $\blacksquare$
The contrapositive is the useful version: if $G$ is non-Abelian, then $G/Z(G)$ is not cyclic. For instance $Z(Q_8) = \{\pm 1\}$, and $Q_8/Z(Q_8) \cong V_4$ (Example 7), which is indeed not cyclic. If $G$ is Abelian, then $Z(G) = G$ and $G/Z(G)$ is trivial, so a nontrivial cyclic $G/Z(G)$ never happens.