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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. Let $H\leq G$. Let $a,b\in G$. Note the $Ha=\{h\cdot a | h\in H\}$. Let $f:H \rightarrow Ha$ be a function defined by $f(h)=h\cdot a$. We claim $f$ is a bijection. We will prove this by showing there is an inverse which we call $f^{-1}$. We claim $f^{-1}(g) = g\cdot a^{_1}$.

We first show $f^{-1}(g) \in H$ for every $g\in Ha$. Let $g\in Ha$. Then pick $h\in H$ such that $g=ha$. We compute \begin{equation} f^{-1}(g)=ga^{-1}=haa^{-1}=he=h\in H. \end{equation}

We now show that $f^{-1}$ is an inverse of $f$. We compute $$\begin{align} f(f^{-1}(g))&=f(ga^{-1})\\ &=f((ha)a^{-1})\\ &=f(h(aa^{-1}))\\ &=f(he)\\ &=f(h)\\ &=ha\\ &=g. \end{align}$$

We also compute $$\begin{align} f^{-1}(f(g))&=f^{-1}(ga)\\ &=(ga)a^{-1}\\ &=g(aa^{-1})\\ &=ge\\ &=g. \end{align}$$ Thus $f^{-1}$ is the inverse of $f$ and we see that $f$ is a bijection.

Now we must prove why we have $|H|=|Ha|$. This is the same as explaining why the coset of $H$ has the same number of elements as $H$. Since $Ha=\{ha|h\in H\}$ we know there are at most the same number of elements, and are left wondering if there are any duplicates. Since we have an injective function $f$ we know there are no duplicates, since each $h\cdot a$ is unique. Thus $|H|=|Ha|$.

Finally, we must prove that we know $b \in Ha$ if and only if $Hb=Ha$. Recall $b\in G$. We suppose $b\in Ha$. Then we pick $h\in H$ such that $b=ha$. We must show both $Hb \subseteq Ha$ and $Ha\subseteq Hb$. First we show $Ha\subseteq Hb$. Pick $x\in Ha$. Then $x=h'a$ for some $h'\in H$. Recall $b=ha$. Since $H\leq G$ we know $a=h^{-1}b$. Thus $x=h'a=h'h^{-1}b$. Since $h'h^{-1}\in H$ we know $x\in Hb$. Thus $Ha\subseteq Hb$.

Now we show $Hb\subseteq Ha$. Pick $c\in Hb$. Then $c=\hbar b$ for some $\hbar \in H$. We compute $c = \hbar b = \hbar ha$. Since $\hbar h \in H$ we know $c \in Ha$. Hence $Ha \subseteq Hb$. Therefore $Ha=Hb$.

We now suppose that $Ha=Hb$ and must prove that $b\in Ha$. Since $H$ is a group we know it has an identity, which we call $e$. Then $e\cdot b\in Hb$, which means $b\in Hb$. Since $Hb=Ha$ we know $b\in Ha$. $\square$

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