H Cxg I

That is, G and H are not invertible if condition (3) does not hold The coefficients a 1 ,a 2 , ··· ,a M and b 1 ,b 2 , ··· ,b N can be determined using Bocher’s.

H cxg i. Hence, we have a onetoone and onto map between the H conjugates of K and the right cosets of N ( K ) ∩ H in H. Therefore, K = h 2 h1 1 Kh 1 h1 2 or h1 1 Kh 1 = h1 2 Kh 2, and the map is an injection It is easy to see that this map is surjective;. =h g d cggV `Y hc \ UjY U b cV `Y d ifd cgY Ub X mYh b ch V Y h\ Y c b Y hc Zi `Z`` h" K Y gYY h\ Uh a Ub m ha Yg k \ Yb k Y hU`_ hc W\ `X fYb " H \ Ym UfY Uk UfY cZ k \ Uh h\ Ym k Ub h hc X c V ih cZhYb h\ Ym WUb b ch e ihY X c h" H \ Ym k `` gUm ¾< YfY `Yh a Y X c h =.

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