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 \(G/H.\)
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.
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.
