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March 27th, 2015, 10:06 PM  #1 
Member Joined: Jun 2012 From: San Antonio, TX Posts: 84 Thanks: 3 Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems  Why does each g in G lie in the coset gH? Where H is a subgroup of G
Textbook: A First Course in Abstract Algebra Rotman I was following a proof of Lagrange's Theorem that states that if $H$ is a subgroup of a finite group G, then $H$ is a divisor of $G$. In the proof it says let $\{ a_1H, a_2H,\ldots, a_tH \}$ be the family of all the distinct cosets of $H$ in $G$. Then $$G=a_1H \cup a_2H \cup \cdots \cup a_tH$$ because each $g\in G$ lies in the coset $gH$ I omit the rest of the proof because I am having trouble seeing why that is so. Take, for example, $G=S_3$ and $H=\left< (1 \ 2) \right>$. $\begin{alignat*}{3} H &= \{ (1), (1 \ 2)\} &&= (1 \ 2)H \\ (1 \ 3)H &= \{ (1 \ 3), (1 \ 2 \ 3) \} &&= (1 \ 2 \ 3)H \\ (2 \ 3)H &= \{ (2 \ 3), (1 \ 3 \ 2) \} &&= (1 \ 3 \ 2)H \end{alignat*}$ Clearly, each $g\in G$ is in $gH$, but why exactly is this so? EDIT: I seem to gain understanding to my question once I type up a question and post it to this forum. My guess is that it has to do with the identity of H. Last edited by MadSoulz; March 27th, 2015 at 10:13 PM. 
March 27th, 2015, 11:03 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  

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