Guest John Doe Posted November 5, 2004 Posted November 5, 2004 I actually spend about 2hrs on this question.Would sombody please help me out. Ques: Let G be a group and H a subGroup og G.Let a,b be elements of G. prove : If a is an element of Hb, then Ha = Hb I Know that I have to show that Ha is a subset of Hb and vice-versa but I am unable to get there
matt grime Posted November 6, 2004 Posted November 6, 2004 if a is in Hb, then there exists some element h in H such that hb=a, Then Ha=Hhb=Hb. since h permutes the elements of H. Of course if you really want to do it the longer way: if x in Ha, then x=ka some k in H, since a=hb for some h, we see x=ka=khb since kh is in H (H is a group and hence closed under composition), it follows that Ha<=Hb Similaly if x in Hb, then x=kb some k, but since a=hb => h^{-1}a=b x=kb=kh^{-1}a which is in Ha since H is a group and closed under inverse and composition.
psi20 Posted November 7, 2004 Posted November 7, 2004 What does it mean by : Let I be a nonempty set of integers that is closed under addition and subtraction. Then I either consists of zero alone or else contains a smallest positive element, in which case I consists of all multiples of its smallest positive element. What's "closed under addition and subtraction"
bloodhound Posted November 7, 2004 Posted November 7, 2004 being closed under addition and subraction means for any two a,b in I a+b is in I and a-b is in I
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