Subgroups, Cosets and Lagrange’s Theorem

Subgroups

Definition. A subgroup $(H, \ast) \le (G, \ast)$ is a subset $H \subseteq G$ such that $H$ is still a group under the operation $\ast$. If $H \le G$ and $H \not = G$, then $H$ is a proper subgroup.

To determine if a subset $H$ of $G$ is a subgroup, we check the following usual subgroup criterion:

Theorem. Let $(G, \ast)$ be a group and let $H \subseteq G$ be a subset. Then $H \le G$ iff

  1. [Identity] $e \in H$

  2. [Closed] for $a, b \in H$, also $a \ast b \in H$

  3. [Inverse] for $a \in H$, also $a^{-1} \in H$

Proof.

$(\Leftarrow)$ Note that associativity in $H$ is inherited from $G$ and these criterion ensures other group axioms for $H$.

$(\Rightarrow)$ Conversely, suppose $(H, \ast)$ is a group. Firstly, let $e_H$ be the identity element in $H$, we have $e_H \ast e_H = e_H$ in $H$, so as in $G$. Thus $e_H = e_H \ast e = e_H \ast (e_H \ast e_H^{-1}) = (e_H \ast e_H) \ast e_H^{-1} = e_H \ast e_H^{-1} = e$. Hence, the identity of $H$ has to be the same as $G$, (1) is true. Secondly, (2) is true as $H$ has to be closed. Finally, as inverse is unique, the inverse of $a$ in $H$ has to be the same as that in $G$ so (3) is true.

The first criteria can be omitted if we know $H$ is non-empty subset. As if $H$ is non-empty, there is some element $a \in H$ and by (3) then (2), $a \ast a^{-1} = e \in H$.

There is also a super-efficient subgroup criterion which is useful sometimes:

Theorem. Let $(G, \ast)$ be a group and let $H \subseteq G$ be a subset. Then $H \le G$ iff

  1. H is non-empty

  2. Given $a, b \in H$, also $a \ast b^{-1} \in H$

Proof.

Similar to the above, $H$ is non-empty implies $e \in H$ (identity).

For all $b \in H$, by (2) and setting $a = e$, we have $e \ast b^{-1} = b^{-1} \in H$ (inverse).

For all $a, b \in H$, we have $b^{-1} \in H$, therefore by (2), $a \ast (b^{-1})^{-1} = a \ast b \in H$ (closed).

The following are some examples of subgroups:

Cosets

Definition. Let $H$ be a subgroup of $G$, i.e. $H \le G$. Given an element $a \in G$, the set

\[aH = \set{ah : h \in H}\]

is a left coset of $H$. Similarily,

\[Ha = \set{ha : h \in H}\]

is a right coset of $H$. (Note: coset is not necessary a group)

Property. Given $H \le G$ and $a \in G$. $a \in H$ iff $aH = H$.

Proof.

$(\Leftarrow)$ If $aH = H$, as $e \in H$, $ae = a \in H$.

$(\Rightarrow)$ If $a \in H$, for any $h \in H$, $ah \in H$ (closure), so $aH \subseteq H$. We also have $a^{-1} \in H$, for any $h \in H$, $a^{-1}h \in H$ (closure), $a(a^{-1}h) = h \in aH$, so $H \subseteq aH$.

Hence, $a \in H \iff aH = H$.

Property. Given $H \le G$ and $a, b \in G$, $aH$ and $bH$ is either the same or disjoint.

Proof.

Assume they are not disjoint, that means for some $h_1 \in H$, $ah_1 \in bH$. Suppose $ah_1 = bh_2 \in bH$, we have $a = bh_2h_1^{-1}$. Thus, for any $h \in H$, $ah = bh_2h_1^{-1}h \in bH$, so $aH \subseteq bH$. Similarily, $b = ah_1h_2^{-1}$, $bH \subseteq aH$. Hence, $aH$ and $bH$ are either the same or disjoint.

Property. $aH = bH$ iff $b^{-1}a \in H$.

Proof.

$(\Leftarrow)$ If $b^{-1}a \in H$, $(ba^{-1})^{-1} = a^{-1}b \in H$ as well. For any $h \in H$,

\[ah = (bb^{-1})ah = b(b^{-1}ah) \in bH\]

similarily $bh = (aa^{-1})bh = a(a^{-1}bh) \in aH$. Hence, $aH = bH$.

$(\Rightarrow)$ If $aH = bH$, say $ah_1 = bh_2$ for some $h_1, h_2 \in H$, $b^{-1}a = h_2h_1^{-1} \in H$ (closure).

Property. All the left cosets of $H$ are of the same size as $H$.

Proof.

Let $aH$ be a left coset of $H$, if $ah_1 = ah_2$, $h_1 = h_2$, hence each $ah$ maps to a different element and $|aH| = |H|$.

Property. Given $H \le G$, $G$ is the union of all the left cosets of $H$.

Proof.

For any $a \in G$, as $e \in H$, $a \in aH$. Hence, the union of all of the left cosets is $G$.

All the above properties is applicable to the right cosets. However,

Property. The number of left cosets and right cosets are the same, i.e.

\[\left|\set{aH : a \in G}\right| = \left|\set{Ha : a \in G}|\right|\]

but in general they partition $G$ differently.

Proof.

Let $H \le G$ and $\phi$ be a mapping from the left cosets to right cosets defined as

\[\forall a \in G, \phi(aH) = Ha^{-1}\]

Assume $aH = bH$, we have

\[aH = bH \implies b^{-1}a = b^{-1}(a^{-1})^{-1} \in H \implies Ha^{-1} = Hb^{-1}\]

Hence, $\phi$ is well-defined.

Suppose $\phi(aH) = \phi(bH)$, we have

\[Ha^{-1} = Hb^{-1} \implies b^{-1}(a^{-1})^{-1} = b^{-1}a \in H \implies aH = bH\]

Hence, $\phi$ is injective.

For all $a \in G$,

\[Ha = H(a^{-1})^{-1} = \phi(a^{-1}H)\]

Hence, $\phi$ is surjective.

In conclusion, $\phi$ is a bijection and therefore the number of left cosets and right cosets are the same.

Left cosets and right cosets in general partition $G$ differently. For example, let $H = \set{e, s} \le D_6$, we have

\[\begin{align*} \set{aH: a \in D_6} &= \set{\set{e, s}, \set{r, rs}, \set{r^2, r^2s}} \\ \set{Ha: a \in D_6} &= \set{\set{e, s}, \set{r, sr = r^2s}, \set{r^2, sr^2 = rs}} \end{align*}\]

For the above theorem, $f(aH) = Ha$ is not a candidate for the proof because it is not well-defined. Consider the same example $H = \set{e, s} \le D_6$, we have

\[\begin{align*} rH = \set{r, rs} &= \set{sr^2s, sr^2} = sr^2H \\ Hr = \set{r, sr} &\not= \set{sr^2, r^2} = Hsr^2 \end{align*}\]

To conclude, the (left/right) cosets of $H$ partition the group $G$, i.e. they are disjoint and their union is the group. Therefore,

Theorem. Let $H \le G$, defining $a \sim b$ if $b^{-1}a \in H$ gives an equivalence relation on $G$, whose equivalence classes are the left cosets of $H$.

Proof.

The conditions for equivalence relation

  • [Reflexive] $a \sim a$ as $a^{-1}a = e \in H$

  • [Symmetric] $a \sim b \implies b \sim a$ as $(b^{-1}a)^{-1} = a^{-1}b \in H$

  • [Transitive] $a \sim b$ and $b \sim c \implies a \sim c$ as $(c^{-1}b)(b^{-1}a) = c^{-1}a \in H$.

Lagrange’s Theorem

Theorem. [Lagrange’s Theorem] If $G$ is a finite group and $H \le G$, then $|H|$ divides $|G|$.

Proof.

Base on the above properties of left cosets, the group $G$ is partitioned by the distinct cosets into disjoint sets with respect to $H$, which is called coset decomposition.

Hence, $|G| = \text{number of distinct cosets} \times |H|$ and $|H|$ divides $|G|$.

Lagrange’s Theorem shows that the order of $G$ influences the number of subgroups $G$ can have, regardless of the group structure.

Definition. The index of $H$ in $G$, denoted by $|G:H|$, is the number of distinct cosets of $H$ in $G$. Hence,

\[|G| = |G:H||H|\]

Corollary. Suppose $G$ is a finite group, for $a \in G$, $\text{ord}(a)$ divides $|G|$.

Proof.

Consider the subgroup $H = \langle a \rangle$ generated by $a$, by Lagrange’s Theorem, $\text{ord}(a) = |H|$ divides $|G|$.

Corollary. Suppose $G$ is a finite group, for $a \in G$, $a^{|G|} = e$.

Proof.

We have $|G| = k \cdot \text{ord}(a)$, hence $a^{|G|} = (a^{\text{ord}(a)})^k = e$.

Definition. The exponent of $G$ is the smallest number $m$ such that $a^m = e$ for all $a \in G$. Thus, $m$ is the least common multiple of all the orders of elements, which are divisors of $|G|$, so $m$ divides $|G|$.

References