关于陪集等价关系$aH=bH \iff H=a^{-1}bH$的推导疑问

Let's break this down step by step—this is a foundational equivalence in coset theory, so it's great you're unpacking the details from Gallian's text. First, recall that $H$ is a subgroup of a group $G$, and $a,b \in G$.

Forward Direction: $aH = bH \implies H = a^{-1}bH$ #

We need to prove two-way inclusion (since set equality requires both subsets to hold):

1. Prove $H \subseteq a^{-1}bH$ #

Take any arbitrary element $h \in H$. By definition of left cosets, $ah \in aH$. Since we're assuming $aH = bH$, this means $ah$ must also be in $bH$. That means there exists some $h' \in H$ such that:
$$ah = bh'$$
Left-multiply both sides by $a^{-1}$ (which exists because $G$ is a group):
$$h = a^{-1}bh'$$
Notice that $a^{-1}bh'$ is exactly an element of $a^{-1}bH$ (since $h' \in H$). So $h \in a^{-1}bH$, which proves $H \subseteq a^{-1}bH$.

2. Prove $a^{-1}bH \subseteq H$ #

Take any arbitrary element $k \in a^{-1}bH$. By definition, this means there exists some $h'' \in H$ such that:
$$k = a^{-1}bh''$$
Multiply both sides on the left by $a$:
$$ak = bh''$$
Since $bh'' \in bH$, and we know $aH = bH$, this means $ak \in aH$. By definition of $aH$, there exists some $h''' \in H$ such that:
$$ak = ah'''$$
Left-multiply both sides by $a^{-1}$:
$$k = h'''$$
Since $h''' \in H$, we've shown $k \in H$. Thus $a^{-1}bH \subseteq H$.

Combining both inclusions gives $H = a^{-1}bH$.

(Note: Your initial draft mentioned $t=ah=bh$—that's a typo, it should be $t=ah_1=bh_2$ for distinct $h_1,h_2 \in H$. Correcting that makes the rest of your derivation align with the above steps: $a^{-1}t = h_1 = a^{-1}bh_2$, which shows $a^{-1}bh_2 \in H$ (supporting $a^{-1}bH \subseteq H$), and the first inclusion follows from reversing that logic.)

Reverse Direction: $H = a^{-1}bH \implies aH = bH$ #

Again, we need two-way inclusion:

1. Prove $aH \subseteq bH$ #

Take any $x \in aH$, so $x = ah$ for some $h \in H$. Since $H = a^{-1}bH$, $h$ can be written as $a^{-1}bh'$ for some $h' \in H$. Substitute this into $x$:
$$x = a(a^{-1}bh') = bh'$$
Since $bh' \in bH$, this shows $x \in bH$, so $aH \subseteq bH$.

2. Prove $bH \subseteq aH$ #

First, note that since $e \in H$ (identity element of the subgroup), $a^{-1}b = a^{-1}b \cdot e \in a^{-1}bH = H$. So $a^{-1}b \in H$, which means $b = a(a^{-1}b) \in aH$. Now, take any $y \in bH$, so $y = bh''$ for some $h'' \in H$. Since $a^{-1}b \in H$, the product $(a^{-1}b)h''$ is also in $H$ (subgroups are closed under multiplication). Let $h''' = (a^{-1}b)h''$, so:
$$y = bh'' = a \cdot (a^{-1}bh'') = ah'''$$
Since $ah''' \in aH$, this shows $y \in aH$, so $bH \subseteq aH$.

Combining both inclusions gives $aH = bH$.

内容的提问来源于stack exchange，提问作者user462561