# Theorem 5, Section 1.4 of Hungerford’s Abstract Algebra

If $$K,H,G$$ are groups with $$K\lt H \lt G$$, then $$[G:K]=[G:H][H:K]$$. If any two of these indices are finite, then so is the third.

Proof: By Corollary 4.3 $$G= \bigcup_{i\in I}Ha_i$$ with $$a_i \in G$$, $$|I|=[G:H]$$ and the cosets $$Ha_i$$ mutually disjoint (that is, $$Ha_i = Ha_j \Leftrightarrow i=j$$). Similarly $$H=\bigcup_{j\in J} Kb_j$$ with $$b_j\in H$$, $$|J|=[H: K]$$ and the cosets $$Kb_j$$ are mutually disjoint. Therefore $$G=\bigcup_{i\in I}Ha_i=\bigcup_{i\in J} (\bigcup_{j\in J}Kb_j)a_i=\bigcup_{(i,j)\in I\times J}Kb_ja_i$$. It suffices to show that the cosets $$Kb_ja_i$$ are mutually disjoint. For then by Corollary 4.3. we must have $$[G:K]=|I\times J|$$, whence $$[G:K]=|I\times J|=|I||J|=[G:H][H:K]$$. If $$Kb_ja_i=Kb_ra_t$$, then $$b_ja_i=kb_ra_t$$ ($$k\in K$$). Since $$b_j,b_r,k\in H$$ we have $$Ha_i = Hb_ja_i = Hkb_ra_t = Ha_t$$; hence $$i = t$$ and $$b_j=kb_r$$. Thus $$Kb_j=Kkb_r=Kb_r$$ and $$j = r$$. Therefore, the cosets $$Kb_ja_i$$ are mutually disjoint. The last statement of the theorem is obvious.

Question: Let $$P=\{Kb_ja_i\mid (i,j)\in I\times J\}$$ and $$Q=\{Kc\mid c\in G\}$$. I think $$G=\bigcup_{(i,j)\in I\times J}Kb_ja_i$$ shows $$P=Q$$ and $$P$$ is pairwise disjoint shows $$|P|=|I\times J|$$.

We claim $$P=Q$$. Clearly $$P\subseteq Q$$. Conversely, let $$Kd\in Q$$ for some $$d\in G$$. Then $$Kd\subseteq G=\bigcup_{(i,j)\in I\times J}Kb_ja_i$$. So $$d=ed\in Kb_qa_p$$, for some $$(p,q)\in I\times J$$. Thus $$Kd\cap Kb_qa_p\neq \emptyset$$. Since two equivalence class are either disjoint or equal, we have $$Kd= Kb_qa_p$$. Hence $$Kd\in P$$ and $$P\supseteq Q$$. So $$P=Q$$. Is my proof correct? It is clear that $$P$$ is partition of $$G$$. Can we use partition argument to prove $$[G:K]=|I\times J|$$? I mean equivalence relation defined by partition $$P$$ is same as right congruence modulo $$K$$.

How to rigioursly prove $$|I\times J|=|I||J|$$? For $$|I\times J|=|I||J|$$ to make sense, we need to define “multiplication” in $$\Bbb{N}\cup \{+\infty \}$$.

Author also didn’t include details of $$(\bigcup_{j\in J}Kb_j)a_i= \bigcup_{j\in J}Kb_ja_i$$ and $$\bigcup_{i\in J}\bigcup_{j\in J}Kb_ja_i=\bigcup_{(i,j)\in I\times J}Kb_ja_i$$ and $$Hb=H$$, for all $$b\in H$$, though these are easy to verify.

$$i):$$ The fact that $$[G:K]=|I\times J|$$ follows directly from $$G=\bigcup_{(i,j)\in I\times J}Kb_ja_i$$ using Corollary $$4.3$$. You don't need a partition argument.
$$ii):$$ Multiplication in $$\mathbb{N}\cup\lbrace\infty\rbrace$$ is simple: when $$a,b\in\mathbb{N}$$, $$a\cdot b$$ is computed as usual, and if $$a=\infty$$ and/or $$b=\infty$$, then we define $$a\cdot b=\infty$$. This is consistent with $$|I\times J|=|I|\cdot |J|$$. If $$I$$ or $$J$$ is infinite, then so is $$I\times J$$. If they are both finite, given $$i_0\in I$$ define $$J_{i_0}=\lbrace (i_0,j)\mid j\in J\rbrace\subset I\times J$$. It is clear that $$|J|=|J_{i_0}|$$. then you have a disjoint finite union $$I\times J = \bigcup_{i\in I}J_i$$, and since it is disjoint, you get the cardinality formula $$|I\times J| = \left|\bigcup_{i\in I}J_i\right| = \prod_{i\in I}|J_i|=|I|\cdot |J|$$. This is basic set theory.
• Thank you so much for the answer. Corollary $4.3$ states “(i) $G$ is the union of the right [resp. left] cosets of $H$ in $G$. (ii) Two right [resp. left] cosets of $H$ in $G$ are either disjoint or equal.“ I agree $[G:K]=|I\times J|$ follows from corollary, to be specific, $\nexists$ $Kc$ such that $Kc\notin P$. I see how you prove $|I\times J|=|I||J|$. Considered cases infinite and finite. Another way to prove it for finite case is as follows: there are $|I|$ ways to choose first component and $|J|$ ways for second component. Thus $|I\times J|=|I||J|$. Commented Apr 28, 2023 at 14:21