AAR Cosets and some applications

Section 2.1 Cosets and some applications

While there are algebraic objects with fewer defining properties than groups (monoids, semigroups, groupoids), the notion of a group is where we start our review of standard results.

One of the most fundamental structure theorems is the theorem of Lagrange.

Theorem 2.1.1. Lagrange’s theorem.

Let \(G\) be a finite group, and \(H\) a subgroup of \(G.\) Then

\begin{equation*} |G| = [G:H]\cdot |H|, \end{equation*}

where \([G:H]\) is the number of cosets in \(G/H.\)

Proof.

Recall that the proof is very straightforward. We know that the (left) cosets of \(H\) in \(G\) form a partition of \(G.\) Since everything is finite let's enumerate the cosets \(G/H = \{ g_kH \mid k=1,\dots, r\},\) so \(r = [G:H].\) Then \(G\) is the disjoint union of the cosets meaning both

\begin{equation*} G= \bigcup_{k=1}^r g_kH \text{ but also } |G| = \sum_{k=1}^r |g_kH|. \end{equation*}

However any two cosets have the same cardinality, \(|H|\text{,}\) so

\begin{equation*} |G|=\sum_{k=1}^r |H| = r|H| = [G:H] |H|. \end{equation*}

Corollary 2.1.2.

Lagrange's theorem not only says that the order of any subgroup divides the order of a group, but also that the order any element in a group divides the order of the group, the later since the order of an element \(x\in G\) equals the order of the cyclic subgroup \(H = \la x\ra\) it generates.

Proposition 2.1.3.

Let \(H,K\) be finite subgroups of a group \(G.\) Then the cardinality of the set \(HK = \{hk\mid h\in H, k\in K\}\) is given by

\begin{equation*} |HK| = \frac{|H| |K|}{|H\cap K|}. \end{equation*}

Proof.

Whether or not \(HK\) is a subgroup of \(G\text{,}\) we can view \(HK\) as a union of cosets:

\begin{equation*} HK = \{hk\mid h\in H, k\in K\} = \bigcup_{h\in H}hK. \end{equation*}

Now all the cosets \(hK\) have the same size as \(K,\) so

\begin{equation*} |HK| = |K|\cdot \text{number of distinct cosets}. \end{equation*}

We see that for \(h_1, h_2 \in H\text{,}\)

\begin{equation*} h_1K = h_2K \iff h_2^{-1}h_1 \in K \iff h_2^{-1}h_1 \in H\cap K\iff h_1 H\cap K = h_2H\cap K. \end{equation*}

Thus the number of distinct cosets in \(\{hK \mid h\in H\}\) equals the number of distinct cosets in \(\{h (H\cap K) \mid h \in H\}.\) But \(H\) is a group and \(H\cap K\) a subgroup, so by Lagrange's theorem, we know that the number of cosets of \(H\cap K\) in \(H\) is

\begin{equation*} |H/H\cap K| = |H|/|H\cap K|. \end{equation*}

Thus

\begin{equation*} |HK| = |K|\cdot \text{number of distinct cosets}= |K| \frac{|H|}{|H\cap K|}. \end{equation*}

Remark 2.1.4.

In the special case that \(HK\) is a subgroup of \(G\) (see Theorem 1.5.3), we shall see this result follows from the second isomorphism theorem for groups.

Example 2.1.5.

Let \(G\) be the symmetric group \(S_3\text{,}\) and let \(H = \la (1\ 2)\ra\) and \(K = \la (2\ 3)\ra\) be cyclic groups of order 2 generated by the given transpositions. It is clear by inspection that \(H\cap K = \{1\}\text{,}\) so \(|HK| = \frac{2\cdot 2}{1}=4.\) By Corollary 2.1.2, since \(4 \nmid 6\text{,}\) \(HK\) cannot be a subgroup of \(G\text{.}\)

Example 2.1.6.

In a different direction, and still with \(G=S_3\text{,}\) if \(H = \la (1\ 2)\ra\) and \(K = \la (1\ 2\ 3)\ra\text{,}\) then \(H\) and \(K\) have orders 2 and 3 respectively. It follows that \(|H\cap K| = 1\) since by Lagrange \(|H\cap K|\) must divide \(|H|=2\) and \(|K| = 3\text{,}\) hence must divide their gcd which is 1. Thus \(HK\) has order 6, so \(HK=G.\) In particular \(HK\) is a group.