Suppose $G$ is a finite group and $H$ a subgroup of $G$. Put $G = \{ g_1,...,g_n \}$ and let $\mathcal{L} = \{ g_1H,...,g_nH \}$ be the set of left of cosets of $H$. Since $H \leq G$, then it is obvious that $|H| \leq |G| $ so $H$ is finite group. IT is obvious also that $| \mathcal{L} | \leq |G|$. How can I show this rigoriously?
Suppose now that $G$ is a group(not necessarily finite). Let $H$ be subgroup of $G$. Let $\mathcal{L} = \{ xH : x \in G \}$ (the collection of all left cosets of $H$) and let $\mathcal{R} = \{ Hx: x \in G \} $. Claim: $| \mathcal{L} | = | \mathcal{R} |$.
Attempt: Define $f: \mathcal{L} \to \mathcal{R} $ by $f(xH) = Hx^{-1} $
suppose $f(xH) = f(yH) $, then $Hx^{-1} = Hy^{-1}$. How can I show that $xH =yH$ ?
For surjectivity, pick any element in $\mathcal{R}$, say $Hx$, we want to find a set $B$ in $\mathcal{L}$ such that $f(B) = Hx$. Choose $B = x^{-1}H$ and we are done. So we obtain surjectivity.
So, I would like to ask how can I prove injectivity and how to prove this function is well defined? thanks
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$\begingroup$For your first question, just note that the map $G\to\mathscr{L}:g\mapsto gH$ is by definition a surjection; the fact that $G$ is finite then ensures that $\mathscr{L}$ is finite.
To show that $f:\mathscr{L}\to\mathscr{R}:xH\mapsto Hx^{-1}$ is well-defined, you must show that if $xH=yH$, then $Hx^{-1}=Hy^{-1}$.
Suppose that $xH=yH$, and let $z\in Hx^{-1}$; we want to show that $z\in Hy^{-1}$. In other words, there is an $h\in H$ such that $z=hx^{-1}$, and we want to show that there is an $h'\in H$ such that $z=h'y^{-1}$. Suppose that we can do this: what will $h'$ have to be? We’ll have $hx^{-1}=z=h'y^{-1}$, so $h'=hx^{-1}y$. Clearly that’s going to be possible if and only if $x^{-1}y\in H$, since it means that $x^{-1}y=h^{-1}h'\in H$. Does the hypothesis that $xH=yH$ let us conclude that $x^{-1}y\in H$? Yes. Indeed, we have the following useful
Fact: For $x,y\in G$, $xH=yH$ iff $x^{-1}y\in H$.
Proof: Suppose that $xH=yH$. $1_G\in H$, so $y=y\cdot1_G\in yH=xH$, and therefore there is an $h\in H$ such that $y=xh$, and $x^{-1}y=h\in H$ as desired. Conversely, if $x^{-1}y=h\in H$, then $y=xh$, and $yH=(xh)H=x(hH)=xH$. (You should check that $hH=H$ for every $h\in H$.) $\dashv$.
Thus, if $xH=yH$, we can let $h_0=x^{-1}y\in H$, and for each $hx^{-1}$ in $Hx^{-1}$ we’ll find that $$hx^{-1}=hx^{-1}(yy^{-1})=h(x^{-1}y)y^{-1}=hh_0y^{-1}=(hh_0)y^{-1}\in Hy^{-1}$$ and hence that $Hx^{-1}\subseteq Hy^{-1}$. Exactly the same argument with $x$ and $y$ interchanged shows that $Hy^{-1}\subseteq Hx^{-1}$, so $Hx^{-1}=Hy^{-1}$, and $f$ is well-defined.
Now see if you can use some of the same ideas to show that $f$ is injective. Your approach of assuming that $f(xH)=f(yH)$ and trying to show that $xH=yH$ is fine, so you’re trying to show that if $Hx^{-1}=Hy^{-1}$, then $xH=yH$; this should look quite a lot like what I just did.
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