# The Number of Elements in a Left (Right) Coset

Recall from the Left and Right Cosets of Subgroups page that if $(G, \cdot)$ is a group, $(H, \cdot)$ is a subgroup, and $g \in G$ then the left coset of $H$ with representative $g$ is defined as:

(1)The right coset of $H$ with representative $g$ is defined as:

(2)We will now look at a rather simple theorem which will tell us that the number of elements in a left (or right coset) will equal to the number of elements in $H$. This seems rather obvious since $gH$ contains elements of the form $gh$ where we range through all of $h$. So $gH$ has at most the same number of elements in $H$. Of course, $gH$ may have less elements if $gh_1 = gh_2$ for distinct $h_1, h_2 \in H$. Of course this cannot be since by cancellation we would then have that $h_1 = h_2$. We make this argument more rigorous below.

Theorem 1: Let $(G, \cdot)$ be a group, $(H, \cdot)$ a subgroup, and let $g \in G$. Then the number of elements in $gH$ equals the number of elements in $H$, i.e., $\mid gH \mid = \mid H \mid$. Similarly, $\mid Hg \mid = \mid H \mid$. |

*We only prove the case of this theorem for left cosets. The case for right cosets is analogous.*

**Proof:**Define a function $f : H \to gH$ for all $h \in H$ by:

- We will show that $f$ is bijective. First we show that $f$ is injective. Let $h_1, h_2 \in H$ and assume that $f(h_1) = f(h_2)$. Then we have that:

- By left cancellation this implies that $h_1 = h_2$ and so $f$ is injective. We now show that $f$ is surjective. Let $gh \in gH$. Then $f(h) = gh$ (somewhat trivially) so $f$ is surjective.

- Since $f$ is bijective we have that $\mid H \mid = \mid gH \mid$ so the number of elements in the left coset $gH$ is equal to the number of elements in $H$. $\blacksquare$