NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/sci/math/11197

Internet Message Format | 1992-09-09 | 2.0 KB

Path: sparky!uunet!mcsun!uknet!gdt!masfeb
From: [e-mail] (F E Burstall)
Newsgroups: sci.math
Subject: Re: Couple of questions
Message-ID: <[e-mail]>
Date: 9 Sep 92 18:23:28 GMT
References: <[e-mail]>
Organization: School of Mathematics, University of Bath, UK
Lines: 41

In article <[e-mail]> [e-mail] (Priscilla V Loanzon) writes:
>Could someone please explain to me a few basic things:
>
>What is the logic used to answer questions of the below type?
>
>1) If the finite group G contains a subgroup of order 7 but no element
>(other than the identity) is its own inverse then the order of group G
>could be (a)27   (b)28   (c)35   (d)37   (e)42.
>
>It says that the correct anwer is (c). I know why we can eliminate (a)
>and (d) but don't know how to proceed further.

Two things are happening here:

1) Good old Lagrange's theorem says the order of the subgrp divides the order
of the grp so the order is a multiple of 7--this, as you know, eliminates (a)
and (d).

2) Since no element is its own inverse except the identity, we can pair off
elements with their inverses and conclude that there is an EVEN number of
non-identity elements and thus an odd number of elements overall.  This
eliminates (b) and (e). Voila!


>Also could someone explain what is elementary Riemann and Lebesgue
>integration, and why fuctions can be integrable but now Riemann
>itegrable etc? I looked over a couple of books but did not quite get
>it.

This is a long story and comes down to the definitions of the different types
of integral.  Given you understand these definitions, here is a quick example
of a non-Riemann integrable function on [0,1]: define f to be 0 on rational
numbers and 1 on irrational numbers.  This fn is not Riemann integrable
(whatever the partition, the upper sum is 1 and the lower sum is 0) but has
Lebesgue integral 1 (since f=1 off a set of measure zero)

>Thanks

Da nada.

Fran