Cosets and Lagrange's Theorem Review

# Cosets and Lagrange's Theorem Review

We will now review some of the recent material regarding cosets and Lagrange's theorem.

- Recall from the
**Left and Right Cosets of Subgroups**page that if $(G, *)$ is a group, $(H, *)$ is a subgroup, and $g \in G$ then the**Left Coset of $H$ with Representative $g$**is defined as:

\begin{align} \quad gH = \{ gh : h \in H \} \end{align}

- The
**Right Coset of $H$ with Representative $g$**is defined as:

\begin{align} \quad Hg = \{ hg : h \in H \} \end{align}

- We then proved a simple theorem which says that if $(G, *)$ is an abelian group then $gH = Hg$ for every $g \in G$.

- We also stated a series of equivalent statements regarding cosets. If $(G, *)$ is a group, $(H, *)$ is a subgroup, and $g_1, g_2 \in G$ then the following statements are equivalent:
- $g_1H = g_2H$.
- $Hg_1^{-1} = Hg_2^{-1}$.
- $g_1H \subseteq g_2H$.
- $g_1 \in g_2H$.
- $g_1^{-1}g_2 \in H$.

- On the
**The Set of Left (Right) Cosets of a Subgroup Partitions the Whole Group**page we proved a nice theorem which tells us that if $(G, *)$ is a group and $(H, *)$ is a subgroup then the set of left cosets, $\{ gH : g \in G \}$ actually partitions $G$. The same statement holds for right cosets.

- On the
**The Number of Elements in a Left (Right) Coset**page we proved that if $(G, *)$ is a group and $(H, *)$ is a subgroup then for any $g \in G$ the number of elements in the cosets $gH$ and $Hg$ is equal to the number of elements in $H$, i.e.:

\begin{align} \quad \mid gH \mid = \mid Hg \mid = \mid H \mid \end{align}

- On the
**The Index of a Subgroup**page we said that if $(G, *)$ is a group and $(H, *)$ is a subgroup then the**Index of $H$ in $G$**denoted $[G : H]$ is defined as the number of left (or right) cosets of $H$. We also proved that the number of left cosets of $H$ always equals the number of right cosets of $H$ so that the index of $H$ in $G$ is well-defined.

- We then proved a very significant theorem. On the
**Lagrange's Theorem**page we proved that if $(G, *)$ is a finite group and $(H, *)$ is a subgroup then the number of elements in $H$ must divide the number of elements in $G$ and moreover:

\begin{align} \quad \mid G \mid = [G : H] \mid H \mid \end{align}

- On the
**Corollaries to Lagrange's Theorem**page we then looked at some nice corollaries to Lagrange's theorem which are summarized below. Let $(G, *)$ be a finite group and let $(H, *)$ be a subgroup. Then:

Theorem |
---|

$\displaystyle{[G : H] = \frac{\mid G \mid}{\mid H \mid}}$. |

For all $g \in G$ the order of $g$ must divide $\mid G \mid$. |

If $\mid G \mid = p$ where $p$ is prime then $(G, *)$ is a cyclic group. |

If $(I,*)$ is also a subgroup of $(G, *)$ with $I \subseteq H \subseteq G$ then $[G : I] = [G : H][H : I]$. |

- We then proved two famous theorems on the
**Euler's Theorem and Fermat's Little Theorem**page. Recall that Euler's theorem states that if $\gcd (a, n) = 1$ and if $\phi (n)$ is the number of positive integers $m$ with $1 \leq m \leq n$ that are relatively prime to $n$ then:

\begin{align} \quad a^{\phi(n)} \equiv 1 \pmod n \end{align}

- As a special case of Euler's theorem we proved Fermat's Little theorem which states that if $p$ is a prime and $\gcd (a, p) = 1$ (i.e., $a$ is not divisible by $p$) then:

\begin{align} \quad a^{p-1} \equiv 1 \pmod p \end{align}