Coset Decompositions and Lagrange's Theorem

Definition 1. Let $H$ be a subgroup of a group $G$, and suppose that the index of $H$ in $G$ is $r$, that is, $[G:H]=r$. Then
$$
G=a_0H\cup a_1H\cup\cdots\cup a_{r-1}H,
$$
where $a_0=e$ and
$$
a_iH\cap a_jH=\varnothing \quad \text{for } i\ne j,
$$
is called a coset decomposition of $G$ with respect to its subgroup $H$.

We now present several applications of coset decompositions and Lagrange’s theorem.

Proposition 1. Let $H$ and $K$ be subgroups of a group $G$, and suppose that $K\le H\le G$. If both $[G:H]$ and $[H:K]$ are finite, then $[G:K]$ is also finite, and
$$
[G:K]=[G:H][H:K].
$$

Proof : By Lagrange’s theorem,
$$
|G|=|H|[G:H]. \tag{1}
$$

$$
|H|=|K|[H:K]. \tag{2}
$$

$$
|G|=|K|[G:K]. \tag{3}
$$

Combining (1) and (2), we obtain
$$
|G|=|K|[G:H][H:K]. \tag{4}
$$

Comparing (3) and (4), we conclude that
$$
[G:K]=[G:H][H:K].
$$
Hence $[G:K]$ is finite.

Proposition 2. Let $H$ and $K$ be two finite subgroups of a group $G$. Then
$$
|HK|=\frac{|H||K|}{|H\cap K|}.
$$

Proof : Let
$$
M=H\cap K.
$$
Since $M$ is a subgroup of $G$, it is also a subgroup of $H$. Let
$$
[H:M]=r.
$$

Suppose that $H$ has the coset decomposition
$$
H=\bigcup_{i=0}^{r-1} h_iM,
$$
where $h_0=e$. Then
$$
HK=\left(\bigcup_{i=0}^{r-1} h_iM\right)K=\bigcup_{i=0}^{r-1} h_iMK.
$$

Since $M\le K$, we have
$$
MK=K.
$$
Therefore,
$$
HK=\bigcup_{i=0}^{r-1} h_iK.
$$

We now show that for all $i,j\in\{0,1,\dots,r-1\}$, if $i\ne j$, then
$$
h_iK\cap h_jK=\varnothing.
$$

Indeed, if $x\in h_iK\cap h_jK$, then there exist $k_1,k_2\in K$ such that
$$
x=h_ik_1=h_jk_2.
$$
Hence
$$
h_j^{-1}h_i=k_2k_1^{-1}\in K.
$$
Also, since $h_i,h_j\in H$, we have
$$
h_j^{-1}h_i\in H.
$$
Thus
$$
h_j^{-1}h_i\in H\cap K=M,
$$
which implies that
$$
h_iM=h_jM,
$$
contradicting $i\ne j$.

Therefore, whenever $i\ne j$, we have
$$
h_iK\cap h_jK=\varnothing.
$$

Moreover, for every $h\in G$, the map $k\mapsto hk$ is a bijection from $K$ onto $hK$, so
$$
|hK|=|K|.
$$

Therefore,
$$
|HK|=\sum_{i=0}^{r-1}|h_iK|
=\sum_{i=0}^{r-1}|K|
=r|K|
=[H:M]|K|
=\frac{|H||K|}{|H\cap K|}.
$$

Proposition 3. Let $H$ and $K$ be subgroups of a group $G$. If both $[G:H]$ and $[G:K]$ are finite, then $[G:H\cap K]$ is also finite.

Proof : Note that for every $g\in G$,
$$
g(H\cap K)=gH\cap gK.
$$

Since there are only finitely many distinct left cosets of $H$ in $G$ and finitely many distinct left cosets of $K$ in $G$, there are at most $[G:H][G:K]$ possible intersections of the form $gH\cap gK$. Hence there are only finitely many distinct left cosets of $H\cap K$ in $G$. Therefore,
$$
[G:H\cap K]<\infty.
$$

Proposition 4. Let $G$ be a group, and let $H$ and $K$ be subgroups of $G$. If $[G:H]$ is finite, then $[K:K\cap H]$ is also finite.

Proof : For every $g\in K$, we have
$$
g(H\cap K)=gH\cap gK=gH\cap K.
$$

Let $[G:H]=r$, and suppose that $G$ has the coset decomposition
$$
G=\bigcup_{i=0}^{r-1} g_iH,
$$
where $g_0=e$. Intersecting both sides with $K$, we obtain
$$
K=G\cap K=\left(\bigcup_{i=0}^{r-1} g_iH\right)\cap K
=\bigcup_{i=0}^{r-1}\bigl(g_iH\cap K\bigr).
$$
For each $i$ such that $g_iH\cap K\ne\varnothing$, choose $k_i\in g_iH\cap K$. Then
$$
g_iH\cap K=k_i(H\cap K).
$$
Hence $K$ is a union of finitely many left cosets of $H\cap K$ in $K$. After removing repetitions, we obtain a coset decomposition of $K$. Therefore,
$$
[K:K\cap H]<\infty.
$$

Proposition 5. Let $H$ and $K$ be finite subgroups of a group $G$. If $|H|$ and $|K|$ are relatively prime, then
$$
H\cap K=\{e\}.
$$

Proof : Let
$$
|H|=m, \quad |K|=n, \quad |H\cap K|=t.
$$
Since $H\cap K$ is a subgroup of both $H$ and $K$, Lagrange’s theorem gives
$$
t\mid m, \quad t\mid n.
$$
Because $(m,n)=1$, it follows that
$$
t=1.
$$

Moreover, $e\in H\cap K$, so we conclude that
$$
H\cap K=\{e\}.
$$

Proposition 6. Let $G$ be a group, and let $H$ be a subgroup of $G$ with finite index $[G:H]=n$. Then for every $a\in G$, there exists an integer $k$ with $0\le k\le n$ such that
$$
a^k\in H.
$$

Proof : Consider the $n+1$ left cosets
$$
H,\ aH,\ a^2H,\ \dots,\ a^nH.
$$
Since there are only $n$ distinct left cosets of $H$ in $G$, at least two of these cosets must be equal. Thus there exist integers $0\le j<i\le n$ such that
$$
a^iH=a^jH.
$$
Multiplying both sides on the left by $a^{-j}$, we obtain
$$
a^{i-j}H=H.
$$
Hence
$$
a^{i-j}\in H.
$$
Taking
$$
k=i-j,
$$
we obtain an integer $k$ satisfying $0\le k\le n$ and $a^k\in H$. This completes the proof.

The cover image of this article was taken at Lake Baikal in Russia.