Cosets

Cosets

Definition (Cosets)

Let \(G\) be a group and \(H \leq G\):

1. A left (right) coset of \(H\) is \(aH = \{ a \ast h:h \in H \}\) (\(Ha = \{ h \ast a:h \in H \}\)), with \(a \in G\).

One usually writes “\(a+H\)” or “\(H+a\)” in additive groups.

Examples (Cosets in \(\mathbb{Z}\) and \(S_3\))

1. Consider \(G = (\mathbb{Z}, +)\) and let \(n \in \mathbb{Z}\). Take \(H = \langle n \rangle = n \mathbb{Z}\). Let \(a \in \mathbb{Z}\). Note that

   \[\begin{aligned} a + H &= \{ a+h:h \in H \} \\ &=\{ a + nk: k \in \mathbb{Z} \} \\ &=\{ t: t \equiv a \quad \mathrm{mod } \hspace{2mm}n \}. \end{aligned}\]

2. Consider \(G = S_{3}\) and \(\tau = \begin{pmatrix}1 & 2\end{pmatrix}\). Take \(H = \langle \tau \rangle = \{ \mathrm{id}, \tau \}\). We have that

   \[\begin{aligned} H \cdot \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} &= \{ \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} \} \\ &=\{ \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 2 & 3 \end{pmatrix} \},\\ \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} \cdot H &= \{ \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 3 & 1 \end{pmatrix} \},\\ H \cdot \begin{pmatrix} 3 & 2 & 1 \end{pmatrix} &= \{ \begin{pmatrix} 3 & 2 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 3 \end{pmatrix} \} \end{aligned}\]

Note that \(H \cdot \begin{pmatrix}3 & 2 & 1\end{pmatrix} \quad \cap \quad H \ast \begin{pmatrix}1 & 2 & 3\end{pmatrix} = \emptyset\).

Lemma (Coset equality criterion)

Let \(H \leq G\) and \(a,b \in G\). Then

\[aH = bH \iff (b^{-1} \ast a) \in H.\]

Proof: (\(\implies\)): We have that \(a \in bH\). Hence there exists \(h \in H\) such that \(a = b \ast h \implies b^{-1} \ast a = h \in H\). (\(\impliedby\)): Suppose that \(b ^{-1} \ast a \in H\). Hence there exists \(h_{0} \in H\) such that \(b ^{-1} \ast a = h_{0}\). Then \(a = b \ast h_{0}\), so \(a \in bH\). Hence \(aH \subseteq bH\), since if \(x \in aH\), we have that

\[x = a \ast h = b \ast\underbrace{ h_{0} \ast h }_{ \in H }\in bH\]

The other inclusion is proved analogously.

Theorem (Any two cosets are equal or disjoint)

Let \(H\leq G\) and let \(a,b \in G\). Then \(aH = bH\) or \(aH \cap bH = \emptyset\).

Proof: Suppose that \(aH \cap bH \neq \emptyset\). Let \(x \in aH \cap bH\). Hence there exist \(h_{1}, h_{2} \in H\) such that \(x= a \ast h_{1} = b \ast h_{2}\). Then \(b^{-1} \ast x = b^{-1} \ast a \ast h_{1} \in H\) , so that \(b ^{-1} \ast a \in H\). By the previous lemma \(aH = bH\)., from which we conclude the result.

Note (Cosets as a partition)

This shows us that the cosets form a partition of the group. Moreover, if we define the relation \(\sim\) on \(G\) by \(a \sim b \iff aH = bH\), this is an equivalence relation.

Definition (Quotient set)

We define the quotient set of \(G\) over \(H\) as \(G/H := \{ aH:a \in G \}\), that is, the set of representatives of \(G\) quotiented by the equivalence class of cosets of \(H\).

Theorem (Equality of the number of left/right cosets)

Let \(H\leq G\). Then the number of right cosets is equal to the number of left cosets.

Proof: Let \(A = \{ aH: a \in G \}\) and \(B = \{ Hb:b \in G \}\). Let \(f:A\to B\) be such that \(f(aH) = Ha^{-1}\). \(f\) is well defined: If \(aH = a'H\), then \(H a^{-1} = H(a')^{-1}\). Finish the proof (exercise).

Proposition (Cardinality of a subgroup and its cosets)

If \(H\leq G\) and \(a \in G\), then \(\lvert a H \rvert = \lvert H \rvert\) (as cardinality, since \(aH\) is not necessarily a group)

Proof: Define \(f:H\to aH\) by \(h \mapsto a \ast h\). Note that

\[\begin{aligned} f(h_{1}) = f(h_{2}) \implies a \ast h_{1} = a \ast h_{2} \implies a^{-1} \ast a \ast h_{1} = a^{-1} \ast a \ast h_{2} \implies h_{1}=h_{2}. \end{aligned}\]

Definition (Indices)

Let \(H \leq G\). We define the index of \(H\) in \(G\) as \([G:H] = \lvert G/H \rvert\) is the number of distinct cosets of \(H\) in \(G\).

Theorem (Lagrange for finite groups)

Let \(G\) be a finite group and \(H\leq G\). Then \(\lvert G \rvert = [G:H] \cdot \lvert H \rvert\). In particular, \(\lvert H \rvert \Big\lvert \lvert G \rvert\).

Proof: Let \(k = [G :H]\). Let \(a_{1}H, a_{2}H, \dots, a_{k}H\) be the distinct cosets of \(H\) in \(G\). Then \(G = \bigcup_{i=1}^{k} a_{i} H\). Since \(a_{1}H, a_{2}H, \dots, a_{k}H\) are disjoint, we have that \(\lvert G \rvert = \sum_{i=1}^{k} \lvert a_{i} H \rvert = k \lvert H \rvert\).

Corollary (The order of an element divides the order of the group)

Let \(a \in G\). Then \(\lvert a \rvert \Big\lvert \lvert G \rvert\).

Proof: Take \(H = \langle a \rangle\). Then \(\lvert H \rvert = \lvert a \rvert\) and the result follows from Lagrange’s theorem.

Corollary (Multiplicativity of the index)

Let \(K,H,G\) be groups such that \(K \leq H\leq G\). Then \([G:K] = [G:H] [H:K]\).

Proof: Applying Lagrange’s theorem to the three indices:

\[[G:H] \cdot \lvert H \rvert = \lvert G \rvert, \quad [H:K] \cdot \lvert K \rvert = \lvert H \rvert, \quad [G:H] \cdot [H : K] = \lvert G \rvert.\]

Thus,

\[[G:H] \cdot [H:K] = \frac{\lvert G \rvert}{\lvert H \rvert } \cdot \frac{\lvert H \rvert}{\lvert K \rvert } = \frac{\lvert G \rvert}{\lvert K \rvert } = [G:K].\]

Proposition (Cosets over the kernel of a homomorphism)

Let \(G, G'\) be groups and \(f:G \to G'\) a homomorphism. Let \(H = \operatorname{Ker}(f) \leq G\). Let \(a \in G\) and \(a' = f(a)\). Then \(aH = \{ x \in G:f(x) = a' \}\).

Proof: Let \(x \in aH\), then there exists \(h \in H\) such that \(x = a \ast h\). Hence \(f(x) = f(a) \ast_{G'} f(h) = f(a) = a'\). Hence \(aH \subseteq \{ x \in G:f(x) = a' \}\). On the other hand, if \(x \in G\) is such that \(f(x) = a'\), we have that

\[f(x) = f(a) \iff \big(f(a)\big)^{-1} \ast_{G'} f(x) = 1_{G'} \iff f(a^{-1} \ast x) = 1_{G'},\]

from which we have that \(a^{-1} \ast x \in H\) and, by the coset equality criterion, \(xH = aH\), whence \(x \in aH\). Thus \(\{ x \in G: f(x) = a' \} \subseteq aH\), and therefore \(aH \subseteq \{ x \in G:f(x) = a' \}\).