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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
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 Originally Posted by MadSoulz 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.
Right. If $e$ is the identity and $H$ is a subgroup then $e \in H$ therefore $g = ge \in gH$. Tags coset, lie, subgroup Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post goodfeeling Abstract Algebra 3 January 19th, 2013 11:24 PM Elladeas Abstract Algebra 2 May 7th, 2011 08:51 AM envision Abstract Algebra 3 October 4th, 2009 10:37 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM signaldoc Abstract Algebra 3 February 19th, 2009 05:50 PM

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