[proofplan] We prove the two assertions by exhibiting inverse maps. Left multiplication by $g$ sends $H$ onto the left coset $gH$, and left multiplication by $g^{-1}$ recovers the original element of $H$. The right coset case is identical in structure, using right multiplication by $g$ and $g^{-1}$. Once the maps are bijections, equality of finite cardinalities follows from the definition of finite cardinality through a bijection. [/proofplan]

[step:Declare the left and right multiplication maps onto the cosets] Define the map \begin{align*} L_g: H &\to gH \\ h &\mapsto gh. \end{align*} This map is well-defined because, for every $h \in H$, the element $gh$ belongs to $gH$ by the definition of the left coset. Define the map \begin{align*} R_g: H &\to Hg \\ h &\mapsto hg. \end{align*} This map is well-defined because, for every $h \in H$, the element $hg$ belongs to $Hg$ by the definition of the right coset. [/step]

[step:Prove that left multiplication by $g$ is a bijection from $H$ to $gH$]We prove that $L_g$ is injective. Let $h_1, h_2 \in H$ and assume $L_g(h_1) = L_g(h_2)$. Then \begin{align*} gh_1 = gh_2. \end{align*} Multiplying on the left by $g^{-1} \in G$ and using associativity and the identity element of $G$, we obtain \begin{align*} h_1 = h_2. \end{align*} Thus $L_g$ is injective. We prove that $L_g$ is surjective. Let $x \in gH$. By the definition of $gH$, there exists $h \in H$ such that $x = gh$. Hence $x = L_g(h)$, so every element of $gH$ lies in the image of $L_g$. Therefore $L_g$ is surjective. Since $L_g$ is both injective and surjective, it is a [bijection](/page/Bijection).[/step]

[guided]We need to show that the map \begin{align*} L_g: H &\to gH \\ h &\mapsto gh \end{align*} is a bijection. A map is a [bijection](/page/Bijection) exactly when it is both injective and surjective. First, we prove injectivity. Let $h_1, h_2 \in H$ and suppose their images under $L_g$ are equal: \begin{align*} L_g(h_1) = L_g(h_2). \end{align*} By the definition of $L_g$, this says \begin{align*} gh_1 = gh_2. \end{align*} Since $G$ is a group and $g \in G$, the inverse $g^{-1} \in G$ exists. Multiplying both sides on the left by $g^{-1}$ gives \begin{align*} g^{-1}(gh_1) = g^{-1}(gh_2). \end{align*} Using associativity of the group operation and the identity relation $g^{-1}g = e$, where $e \in G$ is the identity element, we get \begin{align*} (g^{-1}g)h_1 &= (g^{-1}g)h_2,\\ eh_1 &= eh_2,\\ h_1 &= h_2. \end{align*} Thus different elements of $H$ cannot have the same image, so $L_g$ is injective. Next, we prove surjectivity. Let $x \in gH$. By the definition of the left coset, \begin{align*} gH = \{gh : h \in H\}. \end{align*} Therefore there exists an element $h \in H$ such that \begin{align*} x = gh. \end{align*} But $gh = L_g(h)$ by the definition of $L_g$, so $x$ is in the image of $L_g$. Since $x \in gH$ was arbitrary, $L_g$ is surjective. The map $L_g$ is both injective and surjective, hence $L_g: H \to gH$ is a bijection.[/guided]

[step:Prove that right multiplication by $g$ is a bijection from $H$ to $Hg$]We prove that $R_g$ is injective. Let $h_1, h_2 \in H$ and assume $R_g(h_1) = R_g(h_2)$. Then \begin{align*} h_1g = h_2g. \end{align*} Multiplying on the right by $g^{-1} \in G$ and using associativity and the identity element of $G$, we obtain \begin{align*} h_1 = h_2. \end{align*} Thus $R_g$ is injective. We prove that $R_g$ is surjective. Let $x \in Hg$. By the definition of $Hg$, there exists $h \in H$ such that $x = hg$. Hence $x = R_g(h)$, so every element of $Hg$ lies in the image of $R_g$. Therefore $R_g$ is surjective. Since $R_g$ is both injective and surjective, it is a [bijection](/page/Bijection).[/step]

[guided]We now prove the corresponding statement for the right coset. The map is \begin{align*} R_g: H &\to Hg \\ h &\mapsto hg. \end{align*} Again, it is enough to prove injectivity and surjectivity. For injectivity, let $h_1, h_2 \in H$ and suppose \begin{align*} R_g(h_1) = R_g(h_2). \end{align*} By the definition of $R_g$, this is \begin{align*} h_1g = h_2g. \end{align*} Since $g \in G$ and $G$ is a group, the inverse $g^{-1} \in G$ exists. Multiplying both sides on the right by $g^{-1}$ gives \begin{align*} (h_1g)g^{-1} = (h_2g)g^{-1}. \end{align*} Using associativity and the identity relation $gg^{-1} = e$, where $e \in G$ is the identity element, we obtain \begin{align*} h_1(gg^{-1}) &= h_2(gg^{-1}),\\ h_1e &= h_2e,\\ h_1 &= h_2. \end{align*} Therefore $R_g$ is injective. For surjectivity, let $x \in Hg$. By the definition of the right coset, \begin{align*} Hg = \{hg : h \in H\}. \end{align*} Hence there exists $h \in H$ such that \begin{align*} x = hg. \end{align*} Since $R_g(h) = hg$, we have $x = R_g(h)$. Thus every element of $Hg$ is attained by $R_g$, so $R_g$ is surjective. The map $R_g$ is both injective and surjective, hence $R_g: H \to Hg$ is a bijection.[/guided]

[step:Translate the bijections into equality of finite cardinalities] If $H$ and $gH$ are finite sets, then the bijection $L_g: H \to gH$ implies \begin{align*} |H| = |gH|. \end{align*} If $H$ and $Hg$ are finite sets, then the bijection $R_g: H \to Hg$ implies \begin{align*} |H| = |Hg|. \end{align*} This proves the stated cardinality equalities and completes the proof. [/step]