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Problem 82 (Properties Of Cosets)

Let $G$ be a group, and let $H$ be a subgroup of $G$. Let $a,b \in G$. Prove the following:

  1. The function $f:H\to Ha$ defined by $f(h)=ha$ is a bijection.
  2. We have $|H|=|Ha|$.
  3. We know $b\in Ha$ if and only if $Hb=Ha$.

Solution

Let $G$ be a group, and $H$ be a subgroup of $G$. Let $a,b\in G$.
(1) We will prove $f:H\rightarrow Ha$ defined by $f(h)=ha$ is a bijection.

First, we will show that $f:H\rightarrow Ha$ is injective. This means that if $f(h_1)=f(h_2)$, then $h_1=h_2$ for all $h_1,h_2\in H$.
Let $h_1,h_2\in H$. This means $f(h_1)=h_1a$ and $f(h_2)=h_2a$. Note, since $H$ is a subgroup of $G$ we know that $h_1a,h_2a\in G$.
Now, suppose $f(h_1)=f(h_2)$. This means $h_1a=h_2a$. Now, applying the cancellation law of groups we conclude $h_1=h_2$.

We will now show that $f:H\rightarrow Ha$ is surjective. This mean we must show that for all $c\in Ha$ there exists $h\in H$ such that $f(h)=c$.
Let $c\in Ha$. Now, pick $h\in H$ such that $c=ha$. This means $f(h)=ha=c$.

(2) We will now prove $|H|=|Ha|$.

Recall $f:H\rightarrow Ha$ is injective. Thus, it follows from (1) that $|H|=|Ha|$.

(3) We will now prove that $b\in Ha$ if and only if $Hb=Ha$.

First, let $Hb=Ha$. Pick $c\in Ha$. This means $c\in Hb$. Now, let $h,h^{'}\in H$ such that $c=h^{'}a$ and $c=hb$. This means $h^{'}a=hb$.
Since $h\in H$ then we know that $h^{-1}\in H$. Now by multiplying on the left by $h^{-1}$ we obtain $h^{-1}h^{'}a=h^{-1}hb$. Then by simplifying $h^{-1}h^{'}a=b$.
Now, recall $h,h^{-1}\in H$ and pick $h_3\in H$ such that $h_3=h^{-1}h^{'}$. This means $h_3a=b$ and thus $b\in Ha$.

Now let $b\in Ha$. We will prove that $Hb=Ha$.
Pick $h\in H$ such that $b=ha$. Note since $h\in H$, then $h^{-1}\in H$. Now by multiplying on the left by $h^{-1}$ we obtain $h^{-1}b=h^{-1}(ha)=(h^{-1}h)a=ea=a$. Therefore, we know $Ha=Hh^{-1}b=(Hh^{-1})b$ by CMSPAL. Then by the absorption law $Hb=Ha$.

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