# Thread: Cardinality of right and left cosets.

1. ## Cardinality of right and left cosets.

Suppose that G is a group, H is a subgroup of G, and $\displaystyle g_1,g_2 \in G$. Show that $\displaystyle |g_1 H|=|Hg_2|$.

(Do not assume that G is finite).
I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

Alternatively, is there a way to prove it without using bijections?

2. Originally Posted by Showcase_22
I wanted to try and find a bijection between them but I wasn't sure quite how to do it.

Alternatively, is there a way to prove it without using bijections?

Define: $\displaystyle \phi: g_1 H \to Hg_2$ as $\displaystyle \phi(g_1 h) = hg_2$

Show the map is a bijection.

3. Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

$\displaystyle \phi:g_1H \rightarrow Hg_2$ as $\displaystyle \phi(g_1h)=hg_2$.

So I need to show that an inverse exists ie. $\displaystyle \phi^{-1}(hg_2)=g_1 h$

Is this the best way of doing it or would proving that $\displaystyle g_1H\subset Hg_2$ and $\displaystyle Hg_2 \subset g_1H$ be better?

4. Originally Posted by Showcase_22
Right, i'm not really sure how to show that it's a bijection but here's what i'll try:

$\displaystyle \phi:g_1H \rightarrow Hg_2$ as $\displaystyle \phi(g_1h)=hg_2$.

So I need to show that an inverse exists ie. $\displaystyle \phi^{-1}(hg_2)=g_1 h$
Not really... You have to show that the map is both one-one and onto.

This is equivalent to doing the following steps:

One-One: Show that $\displaystyle \phi(x) = \phi(y) \Rightarrow x = y$

Onto: Show that $\displaystyle \forall y \in Hg_2, \exists x \in g_1H : \phi(x) = y$

Originally Posted by Showcase_22
would proving that $\displaystyle g_1H\subset Hg_2$ and $\displaystyle Hg_2 \subset g_1H$ be better?
This is wrong, be careful.

The question is asking you to prove they have the same number of elements , its not asking you to prove the sets are equal.

The argument you are using is claiming the sets $\displaystyle g_1H$ and $\displaystyle Hg_2$ are the same. But that's not necessarily true here.

5. Define, $\displaystyle \phi_1 : g_1H \to H$ by $\displaystyle g_1h\mapsto h$.
Define, $\displaystyle \phi_2 : H \to Hg_2$ by $\displaystyle h\mapsto hg_2$.
Show that these are bijections (also that the first one is well-defined).
Therefore, $\displaystyle |g_1H| = |H|$ and $\displaystyle |H|=|Hg_2|$, hence, $\displaystyle |g_1H| = |Hg_2|$.

6. Show $\displaystyle |g_1 H|=|Hg_2|$

Injective: $\displaystyle \forall g_1h_1, g_1h_2 \in g_1H, \phi (g_1h_1)=\phi (g_1h_2) \Rightarrow g_1h_1=g_1h_2$

Suppose $\displaystyle \phi(g_1h_1)=\phi(g_1h_2) \Rightarrow g_1h_1=g_1h_2$. (Need To Show: $\displaystyle h_1=h_2$).

Since $\displaystyle g_1h_1,g_1h_2 \in g_1H$ using the property of groups that they have inverses means:

$\displaystyle g_1h_1=g_1h_2$
$\displaystyle g_1^{-1}g_1h_1=g_1^{-1}g_1h_2$
$\displaystyle eh_1=eh_2$
$\displaystyle h_1=h_2$

Therefore the function is injective.

Surjective: $\displaystyle \forall y \in Hg_2 \exists x \in g_1H: \phi(x)=y$

I'm not really sure how to show it is surjective.

I Need To Show: $\displaystyle \phi(g_1h)=hg_2$ holds for all elements in $\displaystyle Hg_2$.

Can I define $\displaystyle \phi$ to be $\displaystyle \phi(g_1h_i)=h_i g_2$ where $\displaystyle h_i \in H$? Since all the $\displaystyle h_i$'s are different this would be a surjection (i'm wondering whether this is a reasonable assumption).

Combining this result with the proof that it is injective would prove $\displaystyle \phi$ is bijective.

7. You are right

Surjectivity is not an assumption.

For the sake of clarity, let me explain the details:

Every element of $\displaystyle Hg_2$ is of the form $\displaystyle hg_2$ for some $\displaystyle h \in H$. Now $\displaystyle \phi$ maps $\displaystyle g_1 h \in g_1H$ to this $\displaystyle hg_2$. Thus every element in the range is mapped onto by some element in the domain and hence the map is onto.